So we have to take up applications of maxima and minima concept today, application of maxima minima. In fact, I'd already started with it in the last class when I discussed with you the nature of the roots of the cubic polynomial equation. So I will say application of maxima minima continued. Okay. So in this segment, we're going to talk about the optimization problems. To be more precise, nonlinear optimization problems. Not like the ones you have in linear programming problem LPP. So in optimization problems, I'll tell you what is the crux of these kind of problems. Here they will be an objective function. Okay. And you have to maximize or minimize that objective function subject to certain nonlinear constraints. So objective function itself may be a nonlinear function of the variable, which is changing. And you also have a constraint, which also is a nonlinear function. So on the basis of this nonlinear constraint, we need to maximize or minimize in short, maximization and minimization. The word itself means optimization and objective function. So we have to maximize or minimize an objective function. So what is the process behind, you know, carrying out this optimization. So let us say this is your constraint function. Okay. So let's say there are two variables involved. In fact, there may be many variables involved. So let's say there is a constraint function like this. Okay. This is the constraint function constraint function. Okay. And there will be an objective function. Okay. So let me, let me just write a generic OX1 for objective function. Okay. So you need to optimize this, optimize this objective function. So what do we do? We first restructure this optimal restructure these objective function in terms of a single variable. Okay. So many times there are multiple variables involved. Let us say I talk about a simple example that let's say there is a rectangle which has got some kind of a length and a breadth. Okay. Length is X, breadth is Y. And I say the perimeter of this rectangle is a fixed value. Okay. That means the length of the, you can say the sides are fixed. But still you want to maximize the area of that particular rectangle figure. So how do you do that? So in this, the constraint is that the perimeter is fixed. You cannot make the perimeter infinitely big. That is the constraint. And the objective function is the area, which is the product of let's say length into breadth. So it's a two-variable function. You need to maximize it. So what do we do in this case? We write one of the variables in terms of the other from the constraint and we convert it to a single variable function. So what do we do? Our effort is to convert it to a single variable function, if possible. Okay. There are many cases where you can manage without also. So we convert it to a single variable function. Okay.  And then in order to have a maxima or a minima for this objective function, the derivative of that function with respect to that variable in which you have written it must be zero. Okay. From here, we get the roots, which are the roots corresponding to the maximum or minimum value of the function. Okay. Let's say the roots come out to be alpha from here. One of the roots is alpha. We basically check whether the double derivative at alpha is positive or negative. If it is positive, that means your objective function f of X is minimum value or has a minimum value at X equal to alpha. And if it is negative, that means double derivative is negative. It means your function has a maximum value at X equal to alpha. Okay. So this is the sequence of steps that we follow. Let's understand it in more details when we take more questions. I'm sure most of you have already done problems based on this and school. So let's get started with some problem solving and we will learn more from the problems themselves, rather than talking about things in general like this. So let me begin with my first question. I wanted to copy this down. I mean you're already aware of the process is just that we have to take up some questions to understand it. So let's start with the very first question. Okay, let's take this question first. This question was asked in ITJ 1991. Normally there is, okay, yeah, that's a very good question. Can there be only one constraint function? If you ask me in general, need or not be, there may be more than one constraint function. Okay. But ultimately, whatever constraint functions you have, you need to write down your objective function in terms of one of the variables. Okay. So yes, there may be multiple constraint functions. But many times we see only one constraint function, two is very rare. All right, so all of you please read this particular question. It says that a window of fixed perimeter, including the base of the arc is in the form of a rectangle surmounted by a semicircle. The semicircular portion is fitted with colored glass while the rectangular portion is fitted with clear glass. The clear glass transmits three times as much light per square meter as the colored glass does. What is the ratio of the side of the rectangle so that the window transmits the maximum light? And sorry, I forgot to wish, wishing you all a very, very festive season of Burma Lakshmi and Muharram as well. These are Indian festivals. I hope you celebrated it during the morning time. Okay. Good. Good to know that. Whereas ours festival is studies. Isn't it? That is the major thing. Okay. So should we discuss this problem? I mean, would you like to try it or should I discuss it with you so that you get an idea how to proceed? How do you want it to go? Okay. You want to try it? Okay. Go ahead. Sure. Could you draw the diagram? Okay. Sure. See, this question has a rectangular structure of the window. Okay. And it is surmounted by it is surmounted by a semicircular part. Let me raise this one. Now, this semicircular is fitted with colored glass. So here you're putting a colored glass. Whereas this part has a clear glass. Okay. Now, what are the restriction here? Restriction is that the perimeter of the perimeter of this particular window is fixed. Okay. The perimeter is fixed. And they're also talking about including the including the base of this particular arc. Okay. So that perimeter is fixed. And you want the light to be maximum from this. Okay. A few things to be taken care of here is that the clear glass transmits three times as much light as the colored glass does. Okay. So the constraint is you may choose your dimensions as per your wins. For example, I may choose this length as X, X, and let's choose this S to Y. So what is given to us is that perimeter, which is 2X. In fact, 2X plus 2X, 4X plus 4Y plus half of this, which is going to be Pi X. This P is a constant. This P is a constant. So this is our constraint. The constraint is this is a constant. And the amount of light which is being passed by this, that is your objective function. Okay. So let us say the amount of light that passes per unit area from the colored glass is lambda. Whatever unit is there for luminous intensity. So lambda per unit, you can say area is the light passing through the colored glass. Okay. So lambda Pi X square by two, that would be the total amount of light coming from the colored part. Okay. And the amount of light coming from the clear part is three lambda times whatever is the area, area will be 2X into 2Y, which is 4XY. So now here I'll stop. Here I will leave up to you to do the working. So you have to maximize L under the constraint P given to you. Could you explain how you got that equation of L? You're talking about L. Whatever. Okay. See, what is the area of this semicircle Pi X square by two? And let's say lambda is some, you can say it is the amount of light passing through that particular area per unit area. Okay. So area into lambda will give you the total luminous, you know, intensity, which is basically crossing that particular part and three lambda because the clear part of this particular window transmits thrice the amount of light. Okay. Normally when we have fancy windows at home, you'll see that people will put two types of glasses. In fact, multiple types of glasses. Okay. So the glass which has been fitted on the top in the semi-circular part that is transmitting three times lesser light as compared to the clear part. So that's why this is lambda and this is three lambda times the area of this particular rectangular part. Okay. Okay. Okay. So I will also know pitching because it might take longer time to solve this. So as you realize that this objective function is in two variables. X is also there. Y is also there. Right. But what we discussed in the previous slide is that we must try to convert our objective function into a single variable function so that we can differentiate that function with respect to that variable. So here I plan to convert everything in terms of X only. So this four Y over here that you see, I will, you know, obtain from the first equation by using P minus pi X minus four X. Okay. That is four Y. And of course there was an X also there. So X will be multiplied to this. If you want, I can multiply X right here itself. Is it fine? Okay. Now for L to be optimum, whether maximum or minimum, I have to, you know, ensure that DL by DX. Okay. Now many people ask me, sir, could I have written everything in terms of Y also? Yes. It is your call. Okay. But write it in terms of one of the variables. In fact, you can use. There's no such rule that you have to always write in terms of one variable. If need be, we can actually work with two variables also. Okay. Yeah. So I will take up a problem based on that type as well. Okay. So now if you differentiate L with respect to X and put it to zero on this side, you will end up getting a lambda pi X. This I'm just splitting it out. It'll give me three lambda P minus six pi lambda X minus 24 lambda X. Okay. Correct me if I'm wrong. And this must be zero. Okay. So from here, from here, you can do one thing. You can drop the lambdas first of all, there's no point of keeping a lambda because lambda will easily get cancelled off. So you can say three P. These two will be collected as minus five pi X and minus 24 X is equal to zero. Okay. So from here, you get your X as three P by five pi plus 24. Okay. So if X is this. Okay. What is going to be your Y. So let us use our constraint given to us. So our constraint says for Y is equal to P minus. P minus pi X. So instead of pi X, I will write as three P by pi plus five pi plus 24. And you have pi X minus four X.  Okay. So this will be your four Y. Let us try to simplify this a bit. I can take a P common and I can take five pi plus 24 common. So it will be five pi plus 24 minus three pi minus 12. Okay. So that is equal to four Y. So four Y is nothing but P times. This is two pi plus 12 by five pi plus 24. Okay. Now what we have been asked, we have been asked for the ratio of the sites of the rectangle so that the window transmits the maximum light. So what I have to find out, I have to find out the ratio, the ratio of the sites, which is let's say two X in this case by two Y. Okay. In short, you have been asked for X by Y. So X is this. Right. Y is going to be. This divided by four. So you can write this divided by four. So you can write P into two pi plus 12. I'm so sorry for writing X here. It is five. Yeah. And five pi plus 24 will be on top, which will get cancelled with this. PP will also get cancelled with this. Correct. And there was a factor of four also, which I forgot. So this is going to be 12 by two pi plus 12. The answer is six by pi plus six. Okay. Now here, one thing I have missed out, which you should not be doing. I have not done the double derivative check. And I've not figured out whether at, at this value of X does the amount of light become maximum or minimum. That test is very, very important. Now I cannot say it is very important for your board levels because even that has become objective. Right. But yes, you must always perform the double derivative test. Double derivative test. In case you are writing the same topic in a subjective paper. So double derivative of L with respect to X calculated at that X, which is three P by a five, five plus 24. This must come out to be negative. Okay. I think it's very easy. You can easily figure it out. So if you do a double derivative here with respect to X, you will end up seeing a lambda pi. This will go off. This will be minus six pi lambda. And this will be minus 24 lambda. Right. And this is going to be minus 24 lambda minus five pi lambda, which is definitely a negative quantity. Assuming my lambda is positive because as the amount of area is increasing, the light passing through will also increase. So this lambda is a positive quantity for sure. Okay. So this test is important. However, this test is not required to be done when you are solving competitive level exams questions, because it will take away probably, you know, 40 seconds or 50 seconds of your time. Okay. So this question came in. IIT J exam. 1991. I think none of you were born with it. By doing that. Is it fine? Any question? Any concerns? How the process is basically, you know, followed. So again, I'll repeat, write down the constraint, write down the objective function, try to write down the objective function in a single variable, differentiate the objective function with respect to a single variable. See, many people say, if I don't convert it to a single variable, then what we can do? What is the other option with you? The other option is let's say you keep it in both X and Y. Then you differentiate it with respect to one of the variables. Let's say X. Then at one point you will see there is a dy by dx coming up. So you can relate dy by dx or you can write dy by dx from the constraint equation. Are you getting my point? So you can replace that dy by dx from the constraint equation. So you can differentiate here both sides with respect to X. Okay. And write down dy by dx in terms of X and replace it back there. That is another way which people follow. It is up to you. Did you understand what did I say? Is the second approach clear? What did I try to say here? See, approach 2 is where let's say you differentiate it L with respect to X. Okay. So what will you get? You will get lambda pi X and here you will get see 12th lambda and you will get X dy by dx plus Y, isn't it? Now you differentiate this side also with respect to X. So dp by dx will be 0 and this will be 4 and this will be 4 dy by dx. Okay. So this dy by dx, you can find it out and replace it over here. That's another way to do it. Okay. Depending upon what is the need of the hour, you can follow this approach also. Is it fine? So I'm not doing it. Maybe you can try it out on your own as well. Okay. Let's move on to the next question. We will take few questions to understand the nitty gritties of the situation. Okay. Let's take this question, a simpler one. A right circular cone of maximum volume, which can be inscribed in a sphere of radius R has altitude equal to which of the following options. Once you're done, give me a response on the pole. I will be running the pole as well, but not now, maybe in few minutes time. Once you're right, meanwhile, as you're solving it, I will make the diagram. So this is the sphere under which you have to inscribe a right circular cone. Okay. Right circular cone is a cone where the axis of the cone is perpendicular to the base. Now, why can't this right circular cone have infinite volume? If you want to maximize volume, it can go to infinity, right? The problem is that it has a constraint. The constraint is it has to be within a sphere of radius R. That is the constraint, right? It's like in our day to day life, we have constraints related to so many things. We have constraints with respect to time. We only have a 24 hours day, right? So God has not given us infinite long day. So we have to make use of, you know, that constraint and maximize our objective. Okay. So this particular right circular cone has a constraint that it cannot go beyond this particular sphere. And it cannot be of any orbit shape. It has to be a right circular cone. Okay. So, let us first wait for people to give a response. Okay. Shrita has given. So let me put the poll on because... Oh, sorry. Poll has not been created. Anyways, you can give me a response on the chat box. Okay, Gayatri. Okay. I got that. I could figure out that symbol was right above 3 actually. Aditya. Okay. So two different answers so far. Okay, Pratik. Above. Okay. Let's discuss it. Now, many times we would require some extra, you know, a perimeter to help us solve the question right now. I believe that if I could take this angle as theta, my life will be simpler, right? So I'm not claiming that you cannot solve the question without it. But taking this as theta will really help you out. Okay. Now, here we have... We can directly write down the objective function. So our objective function is basically to maximize the volume of this cone. So let us write down the volume of this cone in terms of the radius of this particular sphere and this angle. Now, I can clearly see that the radius of this particular cone is given by... This is r. This is theta. So this is r sine theta. So it is r sine theta whole square.  Okay.  So you can see here that your objective function is automatically written in terms of one variable. And here the variable is theta. So theta is the only variable because if you change the dimension of this right circular cone, theta is going to vary. Nothing else is going to change. So let me write it in a... Yeah. In a sanitized manner. Now here, the derivative of the volume with respect to theta must be zero if you want your volume to be maximum or minimum. Okay. So this condition is the necessary condition for both maximum and minimum. So this particular condition must be satisfied. So when you differentiate it with respect to theta, you will end up getting... This is nothing but sine square into minus sine, which is minus sine cube. And this is going to be one plus cos theta into two sine theta cos theta. Okay. This should become a zero. Okay. So from here you can say that sine cube theta is equal to two sine theta cos theta one plus cos theta. Okay. Now remember sine theta cannot be zero because if theta is zero, this particular right circular cone will collapse. Okay. Neither can it be pi. Pi also will make it collapse. Okay. So one of the sine theta's get cancelled off. Okay. And you have something like this. You can write your sine square as one minus cos square and one minus cos square is one plus cos theta, one minus cos theta. And you can cancel off this as well because you know cos theta is not minus one because for that theta will become pi, which is not possible. So this gives us three cos theta is equal to one. So cos theta is one third. Okay. Now please do our double derivative tests, which I'm not doing here. The double derivative of the volume with respect to theta at cos inverse one third, this should come out to be negative. Okay. So please perform this test. Okay. Internally. And now they're asking us for the height.   Option number C is correct. Option number C is correct. I think most of you got it right. Apart from. Aditya, what happened? You're the only person to get this wrong. Yeah, please check your, please check your working. Okay. Next question. Okay. So we'll take some. So this question says an electric light is placed directly over the center of a circular plot. Of lawn 100 meters in diameter. Assuming that the intensity of the light varies. Directly as the sign of the angle at which it strikes an illuminated surface. And inversely as the square of its distance. From its surface. Okay. How should the light be hung in order that the intensity may be as great as possible at the circumference of the plot. Read this question. Understand this question. What it is trying to say. Oh, sorry. Distance is from the light source. I mean, you would have done it in physics as well. Right. So the intensity of a light. Varies inversely square. Of the distance from the light source. Correct. So let's say this is the spot on the circumference. Let's say there's a stadium. There's a running track. Now you want the athletes to practice during the night. Okay. And they're running on the track. So at whatever spot the athlete is on the track, the distance of that. Distance of that point from the light source. The intensity varies as inversely square of that distance. Getting my point. So this distance is what you are looking for. This is your distance. Okay. And it varies as the sign of this angle. Correct. So the farther away the man from the light source, obviously distance is also increasing and sign is reducing. So the intensity of light will keep on diminishing. It's a very easy question.  That's what that is the purpose. No. You want it to be as great as possible on the circumference. So every point will be at the same distance. That is why you are hanging your hanging the line in the, in the center of this lawn. See the question is directly over the center of the circular plot. Okay. So wherever you take your S. Right. All the points will get the same intensity. So you can just worry about one point being maximum. All the points will be taken care of automatically. Yes. Should we discuss it now? See, very simple. This is 50. This distance is 50 from here to here. Okay. So your D is going to be 50. Seek theta and the amount of light. Let us say the intensity of light that is falling is going to be proportional to let me write proportionality constant. Propositional to sign of the angle divided by square of the distance, which is 50 C theta square. Okay.  Okay. So now you want your intensity to be maximum. So if I has to be maximum slash minimum, it implies that di by d theta must be zero. Correct. So first of all, writing this properly, it will be a lambda by 50 square. And this is going to be sin theta cos square theta. So di by d theta lambda by 50 square. You can write as it is. And derivative of this is going to be a cos cube theta minus sin theta and minus two sin theta cos theta. So it's minus two. And this must be equal to zero. This must be equal to zero, which clearly implies that cos cube theta should be equal to two sin square theta cos theta. You know, cos theta. Cos theta cannot be zero because for that theta has to be 90 degree, which cannot happen. Okay. Because for that this, the track has to be of zero radius. So cos square theta is equal to two sin square theta. In short, you can get your tan theta as one by root two. Okay. So if one by root two is this, what is the height? What is this height at which you should have hung this light source? The light is nothing but at a distance of 50 tan theta. Isn't it? So your answer will be 50 by root two. That's 25 root two meters. So the person or the stadium management should hang this light at a distance of 25 root two meters from the, from that track, you know, in the, up in the air. Okay. And again, I have not done the double derivative test, but if at all there is a subjective paper, you must check the double derivative at tan inverse one by root two to check whether this is negative or not. Please do this checking. I am not doing it. In the interest of time. Okay. It's just a manual work. Is it fine? Any questions? Any questions? Any concerns? Please note down anything. If you want to ask anything, please do so. Else you'll move on to the next question. Clear? Can I move on to the next one? Okay. I think they are missed out a minus. Next question is a geometry based question. So there are three points given to you. A, B, C, A is P square minus P. B is Q square minus Q and C is R square minus R. They are the vertices of a triangle ABC. A parallel gram, A F D E is drawn with vertices. D, E, F on the line segments, B, C, C and A, B, respectively. So, A is P square minus P, B is Q square minus Q and C is R square minus R. They are the vertices of a triangle ABC. They are the line segments, B, C, C and A, B, respectively. Show that the maximum area of such a parallel gram is this. Okay. Ideally they should write a modulus here. I mean, it's a good practice to write a model. Yeah. So read the question once again. Please ensure you have understood it well. In fact, I'll draw a diagram also for you. Okay. And there's a parallelogram A F D E, which is drawn with D on BC. So let's say D is here somewhere. E on C A, let's say E is here somewhere. In fact, I have to make a parallelogram, right? So I must ensure D is parallel to AB. Okay. And F is somewhere over here because I have to ensure DF is parallel to AC. This parallelogram will be constructed. So this is a parallelogram. This is parallel to this and this is parallel to this. You want the area of the parallelogram A F D E to be maximum. Okay. Let us try to solve this as it'll take a lot of time. See, let us write down some lens over here. Let's say I call this lend as X. Okay. And I call this lent as let's say Y. Okay. And this angle is angle A. Okay. So in terms of XY and A, what do you think is the area of the parallelogram? Just type that in your chat box. In terms of XY and angle A, what is the area of the parallelogram? XY sine A. Correct. Everybody agrees with it? Correct. In this objective function, what is constant A? Because the vertices of the triangle is fixed. What is changing X and Y? Okay. So you can have different types of parallelogram, which I, which can be made, but you have to make a parallelogram with the maximum area. Okay. So there are two variables X and Y and there's one constant A. So in order to differentiate it first, I need to convert it to a single variable objective function. Correct. How will I do that? In class 10s, you would have studied Thales theorem, basic proportionality theorem. That is what we call it. Right. So if the two sites are parallel in a triangle, let's say you draw it, then we can say that the ratio of the sites will be same as, let me write it down. So can I say in this triangle CDE and triangle CBA are similar? Right. So can I say this, that DE, which is also X. Okay. DE is also X. Please note this down. So DE by, let's say AB, that is your length, let's say small c. We normally call the side opposite to angle C as small c in a triangle. Okay. This is equal to, this is equal to, can we say CE by CA? Okay. And CA is what? CA is small b. Let me write it down here. Okay. Side opposite to. So basically I'm not touching the coordinates as of now. Many people will say, where are you using the coordinates? As of now I'm not using the coordinates. I'm just basically using my side lengths of the triangle to get my job done as far as I can.  Why they will be constants is because the vertices are fixed. So side lengths will also be fixed. Isn't it? So why am I doing it? I'm doing this to get a relationship between X and Y. Are you getting my point? What is the rationale behind it? To get a relationship between X and Y. So DE is X. Right. And AB is C. CE. CE will be, if I'm not mistaken, Y minus B. By CA, CA will be B. In short, you end up getting, correct me if I'm wrong. X by C is one minus Y by B. So Y is equal to B times one minus X by C. Okay. So this I will substitute over here. This is just to make my objective function P as a single variable function. So I'll get X. Y will be B one minus X by C sine A. Okay. In short, it will become B sine A. And this will be X minus X square by C. Now, in order to make your P maximum or minimum, the derivative of P with respect to X must be zero, which means B sine A, one minus two X by C must be zero, which clearly means this must be zero, which means X must be C by two. Correct. Convinced. If X is C by two, you will check that Y becomes, if you put C by two here, Y will become B by two. In short, the area of the parallelogram, in short, the area of the parallelogram becomes maximum when X is C by two, Y is B by two. And of course, sine A will remain sine A. In short, in short, it's half of, half of B C sine A. Who will tell me what is half B C sine A? What does this represent actually? What is half B C sine A? Right. It's the area of the triangle A B C. Now here, the role of coordinates will come into picture. Okay. Now, how do you find the area of a triangle when you know its vertices? Everybody knows the formula. Area of a triangle, when the vertices are known is half X one, Y one one, X two, Y two one, X three, Y three one. Correct. And most of you have done determinants also. I didn't get that. Could you try again? Okay, Siri.  I don't know why my computer wants to participate in the class. I don't know. Okay. So what is X one p square? If I'm, if I could just refer to the coordinates once again, give me one minute.  If you have all done determinants in your school, can I use a property of determinant by which we can take one row and subtract it from the others? Have you done that property in school in determinants? Yeah, I'm sure you must have done. Rashmika. Thank you. Thank you for confirming that.  So I'm taking my third row and properties is deleted, but still these, these concepts you must have done. What among us hasn't done it? Oh, you're Rama. Aditya P. Oh, Aditya means DPS, right? Oh, HSR has done it. Okay. What should I do? Okay. Let me teach you something here before I solve it. What about gear international? Gaurav.  Gear has done it. Gear SSRVM, et cetera. Believe it or not. Fast, fast, fast. Okay. Gaurav has done it. SSRVM. Anybody from SSRVM? Aditya is there. No, Aditya is not there. Easy. Let me teach you this. This is not a rocket science. See, there is something called row transformations. If you take any row and subtract it from the other rows, then this doesn't change the determinant. Okay. Again, there is a story behind it, which I will take when we start with determinist topic, which immediately after this we are going to do, but not today. In fact, today we are just going to start with matrices. Okay. So anyways, the answer here would be half. If I do this operation, it will be P square minus R square. This will become R minus P zero. Q square minus R square. R minus Q zero. R square minus R and one. Okay. And you can take your P minus R, etc. common out from first row.    And now you can expand it. You can expand it with respect to column number three. So if you expand it with respect to column number three, this is what you end up getting. Yeah. Negative of this plus this.  So minus P minus R. So it'll be just getting Q minus P from it. By the way, in order to keep your answer positive, you must take a modulus of this entire thing. So it'll be half modulus P minus R, Q minus R and Q minus P. So this is the area of the triangle. Okay. What I've been asked to us is the area of the parallelogram, which we figured is going to be half of this, right? So this answer is going to be, I'm writing it in different color. It's going to be half of half, which is one fourth of modulus P minus R, Q minus R. P minus R, Q minus R and Q minus P. In fact, this is what we had been asked to prove. This is what we had been asked to prove. Yeah. Please ignore the order in which they have written because it has been modded.  So this is how you solve this question. So here you can see that the coordinates were involved at the end, not in the, in the beginning. Is it fine? Any question with respect to the way I solved this question? So I wrote the area of the parallelogram in terms of X, Y and sign of the angle A, then I got a relationship between X and Y by using my basic proportional theorem. Once I got it, I differentiated it, differentiated the objective function, put it to zero. I got my X and Y, which I realized that we are such that it gave you the maximum area of the parallelogram as half the area of the triangle. Then I use the coordinate to find the area of the triangle and I halved it and I got the result. That we'll discuss when I do determinants, Aditya. In fact, if you have one, one, one in, you know, any one of the rows or a column, you can take any one of them to subtract with the other two.  Take any one of them. Take R1 subtracted from R2 R3. That will not make a difference to your answer. Okay. Any questions? A regular base, a regular square-based pyramid is inscribed in a sphere of given radius R so that all the vertices of the pyramid belong to this sphere, means lie on this sphere. Okay. Find the greatest volume of the pyramid. I hope everybody knows the fact that volume of a pyramid is one third area of the base times height of the pyramid. And this is applicable even to a cone. Cone is also a pyramid, right? So half area of the base in a cone is pi R square. Height is H. So this is valid even for a cone. So in this case, your pyramid has a square base and it has been inscribed in a sphere of radius R. So they're asking us for the greatest volume of the pyramid. Let me make a diagram. Basically, I'm showing the periphery of the cone. Drawing may not be very good. So please bear with me. So basically, I've shown that these four square base, this is your square base. Let's say I call it as ABCD. They're hitting against the surface of the sphere. Okay. So the part where they're hitting the surface of the sphere, I have shown it with this dotted circle. Okay. And this is the other vertex. You may call it as point, let's say. Oh, awesome. Very good. Very good. Let's say this is our center of the sphere. So far, Aditya has given me a response. Let us see whether that is right or wrong. Okay. All of you, please pay attention. I'm going to slice this sphere along with the pyramid along the taking a plane which passes through E and the diagonal AC. In short, my plane is like this. So this is a plane. This is a plane which is cutting the sphere and it's passing through the vertex E and the diagonal off and the diagonal off the square base. I hope you can imagine that. So I took a plane and I sliced the sphere in such a way that my plane is passing through E and the diagonal of the sphere. Okay. Now, why did I do that is because I wanted a triangular structure like this to work on my, so let's say this is your distance. This is your radius of the, this thing, sphere. Okay. So now if you see this particular cross-sectional view, so if you see from this direction, so you realize that you are going to see something very similar to what you saw when we are dealing with the cone question. A little while ago we took a cone question, right? So the way you dealt with the cone question, the scenario will become more or less similar, but of course the objective function is going to vary a bit. I made this thing in gray color, so I'll continue with the gray color only. Let me choose a triangular structure only. Okay. Now please understand here that this length that you are seeing is the length of the diagonal of this particular right circular cone, diagonal of the base of the pyramid. Are you getting my point? Okay. So let's say this is your R. This is also your R. This is also your R. Let me drop a perpendicular. This will be let's say an angle theta. Okay. And this is your AC. These two vertices AC and E is on the top over here.  Understand the figure. Then only I will proceed further. There's no point proceeding further without understanding the fact. Clear. Any questions? Okay. So you realize that EM would be the height of the, the height of the pyramid pyramid would be EM. Okay. And that happens to be capital R. And this length is our cost theta. So capital R plus our cost theta that happens to be the length of the pyramid. Okay. Now what is MC? MC is half the diagonal. Right. So MC is R sine theta, which is half the diagonal, which is half the diagonal of the square base. Are you getting it? So this, when you drop down, it will drop here at this point M will be here. So this is your half the diagonal. I hope you are able to relate to this figure. Both these figures you are able to relate. One, of course, is a 3D figure. I have tried to make one. And this is a 2D figure. So MC is half the diagonal square, half the length of the diagonal of the square base. So diagonal itself is to R sine theta. Okay. Now, normally let's say if this length is small a. Okay. Then root to a is your length of the diagonal. So let a be the square side length. Okay. So this is a, a, a, a. So root to a is 2R sine theta. So a will be root 2R sine theta. Okay. Now, having known the side length, what is the area of the base? Area of the base will be square of this, which is 2R square sine square theta. So what will be the volume of the pyramid? Volume of the pyramid is one third area of the base into length. Right. Yes or no. So dear all, this is very similar to the expression, which we saw a little while back.  So if this has to be zero. It means cost theta is one third. Check your result. We had done it, you know, in the, in today's session itself, a little while back. Okay. So once cost theta is known to you volume of the, I'm so sorry. Volume of the pyramid will be 2 third R cube sine square theta will be one minus one by nine and this will be one plus one by three. So your answer will be two third. This is going to be eight ninth. This is going to be four by three R cube. How much is this 16 into four 64 by 81 R cube, 64 by 81 R cube. And please also check that the double derivative of this with respect to theta will be negative. You can please make a check out of it. I will not do it. And I think let me see what was your answer. Aditya, absolutely correct. Well done. Aditya, did you find out an easier way to do it? Because you give the answer very fast. Any approach that you would like to share with the class that you took? I took theta the other way. I mean, the approach means did you, did you take the cross section itself, et cetera? I mean, if you could unmute and talk. Yes, yes. We can hear you. Aditya. I took theta the other way. I thought it would be easier to work. From university I thought it would be easier. But the approach was the same. Okay. So you took this angle as theta here. This guy is theta. Yes, sir. Yes, sir. Thank you. Thanks, Aditya. Is this fine? Any questions? Any concerns? By the way, I have not done the derivative and solve for theta, but since this expression was already known to us, I utilized that and saved my time. Is it okay? Could you repeat the path after differentiation? See, after I differentiated it, which I did not show here is it done because I had already done this part when I was dealing with the cone question. I think you were there in the session, right? So just check that part. We had derived that cos theta will become one third. So I use that directly over here to get the maximum volume. I use that. So sine square is one minus cos square. So one minus one third square and one plus one by three. Simplification. I did. That's it. Is it clear? Any questions? Okay. Let's move on to next question. Yes, we'll take this question. This came in ITJ 2015. Okay. Read this question. Everybody. I'm using my artistic skills here. So basically it is a question where you have been given that there is a cylinder. You can say there's a flask kind of a thing, which has got a fixed inner volume. Okay. So this volume is fixed. Okay. This is fixed as a V meter cube. V millimeter cube. And it is two millimeter thick wall. So this, this thickness is off two millimeters. And it has got a disc. This is a disc at the bottom. This thickness is also two millimeters. Now read the question carefully. The question says, if the volume of the material used to make the container is minimum, when the inner radius of the container is 10 mm, find the value of V by 250 pi. Now up till now what used to happen, it used to happen that we used to find out the value of the variable for which the objective function is maximum. Right. Here, the question has been designed in a slightly reverse manner. They have said that it is maximum when the variable value is 10. And they're asking you for the fixed quantity values. So V is actually a fixed value. As you can see, they've already said it has got a fixed inner volume. So the question has been designed in a reverse psychological manner. So instead of asking you for the value of the inner radius, which in this case is a variable, they have said that when the inner radius is 10, then the volume required to make the material, not the volume inside. So whatever material is required to complete the flask, that becomes minimum. Okay. By the way, I'll just remove this. This is just confusing the entire figure. So there is a, yeah, there is a solid disk down over here. Okay. So in order to make the volume of the material, I'll write it as VM to just distinguish you to be minimum. This R must be 10. So what should be the volume? What should be this fixed volume? So see the psychology in which they have played on. Right. So a person who has only practiced the other way round type of problem may find it slightly weird to read. That also, I mean, till now I was always solving for the unknown. Here they have given me the unknown. And I always had a known and they're, they're, they're asking you the known value or the fixed value. I hope I have been able to explain you the scenario here. Yes. Any progress? Okay. Okay. Good. Okay. Okay. Let us write down the expression for the volume of the material required. So volume of the material required will be nothing but first of all, volume of the disk down below whose area, whose volume will be Pi. Now let's say inner radius is R, but the disk radius will be R plus two actually. So Pi R plus two square into height, height is two. Okay. So this is the volume of the disk, which is down at the bottom of the flask. Now what about this annular cylinder? Annular cylinder will be having an area of Pi R plus two, the whole square H minus the volume inside. Isn't it? Let's say H is the height. Let's say H is the height of this particular hollow part inside the cylinder. Right. So Pi R plus two square H will be the outer part. And from inside, you remove that volume V that will be the volume of the material required. Isn't it? Now here there are two variables. One is H and other is R. Okay. So what I'm going to do is by the way, this V is a fixed value. It's a constant for us. So what I'm going to do is I'm going to write down my H in terms of R, which is very easy to write because V itself is Pi R square H. So from here, you can say H is V divided by Pi R square. Okay. So I'm going to replace that over here. Before that we can take, I think Pi R plus two whole square two plus H. Okay. Minus V. And this H. I'm going to replace with V by Pi R square. Okay. This is the volume of the material required. Volume of the material required. Is it fine? Any questions? Any concerns so far? Okay. Let me just do a quick simplification here. I will take R plus two whole square inside. Plus volume. Pi Pi will get canceled. So you have R plus two whole square by R square. Sorry. This, this will, this is already taken care of. So there will be a Pi. So far so good. Okay. So let's now differentiate this with respect to R and put it to zero. So Pi, this will be four R plus two. And this will be V by Pi. I can treat this as if you have written R plus two by R, the whole square, or you can say one plus two by R the whole square. So derivative of that will be two one plus two by R into derivative of this part will be minus two by R and this volume will be differentiated to zero. Okay. So from here, we end up getting four Pi. R plus two is equal to two V. One plus two R into two by R square. And this equation is satisfied. This is satisfied by R equal to 10 as per our question. See. Read the last statement here. If the volume of the material uses minimum when the radius is 10. So when R is 10, this is minimum. That means this condition is satisfied. So you put your value of, you put your value of 10 in your expression. So this will become 12. This will become one plus one by five into two by 100. So let's simplify this. So four Pi into 12. This is going to be two V. Six by five into two by 100. So I think this will cancel off the 12 part and this will cancel off the two part. Okay. So volume becomes 1000 Pi. Okay. Check it out. So in the question, they're asking you the value of V by 250 Pi because they wanted to make an integer type question out of it. So this divided by 250 Pi will give me an answer of four. Is this fine? So the reason I took this question was because I wanted to show you a different perspective of how a question can be framed. Okay. So with this, we conclude this chapter. I think you have already been given a worksheet on Maxima Minima, which has got word problems there. Okay. You can try that out for your assignments and now I'm going to begin with a new concept.


The last thing that Dr. Marek said to me as I was coming up here to introduce her as well, by moving down to the front row at least there's one thing less to trip over. Well, I've already proved that it's still possible to trip over things. However, it is a very great privilege and a great personal pleasure to introduce Dr. Marek. We have somewhat similar backgrounds. We both left a cold rainy England, she from Cambridge to go to San Diego, and I came to North Carolina when I first came over here. And our paths crossed again in Rochester, New York, and I have to of course add New York to that title here in Minnesota where she was on the faculty at the University of Rochester when I first joined the faculty at the University of Rochester. So it's a particular pleasure for me to be able to introduce Dr. Marek. From Rochester in 1979, if I remember the date correctly, she went to Colorado to the University of Colorado Health Science Center to take up a position there, and she has been an investigator of the Howard Hughes Medical Institute there since 1986. Those of you that have read her biography in the brochure will know that it would take up most of her time speaking if I go through that. So I will just mention a few of the highlights of her achievements, perhaps foremost of which is the fact that she was elected several years ago to the National Academy of Science at a time when both young people and young women were not being elected regularly to the National Academy of Science, and many of us are very proud of the fact that she was elected to the National Academy. Her talk this afternoon emphasizes much of the work that she has done over the years, and she has been, you heard Dr. Manasara this morning, while Dr. Marek has taken another part of the immune system and made it understandable at a level that 10 years ago or 15 years ago we just could not have dreamed about, and as a biochemist standing here it's a particular delight to know that she's going to talk about the T-cell receptor and some of its biochemistry and immunology. So with no further ado I will introduce Dr. Phillip Marek and the title of her talk, T-cells and Health and Disease. Dr. Marek. It's really very wonderful to get the opportunity to talk to you all today. I think for two reasons, well for one reason, and that is that as a basic scientist there are two things that, and I expect to those of you who are scientists too know this, that there are two things that really give you pleasure in life. One of them is to go in the lab and look at this big machine and it's spitting out the results, and all of a sudden you realize that you know something from the results that you never had any idea that it existed before, so you discovered something new, that's really very exciting, and then the next thing that's really, the next thing that you do under those circumstances and it's the next thing that gives pleasure is to rush out and tell everybody else what it was that you found so they can all share in this wonderful discovery that you've made. So I hope that some of you will follow along with what I'm talking about today and get some pleasure out of the beauty of the science that I'm going to talk about. Not because we discovered it, but because it was there already to be discovered and already very beautifully arranged and set up so that it functioned very well. And what I'd like to do today is to follow along some of the points that Dr. Banasra have already introduced you to this morning. In particular, this idea that one of the things the immune system has to do in each of us is detect that foreign things have come into the body so the immune system has to be able to recognize just about all the foreign things that could enter your body. It has to have foreknowledge of what those things might be or has to anticipate them. On the other hand, simultaneously, the immune system must not be able to recognize any of the things that are in the body already, any of self. So the immune system must tell the difference between everything else and you. And if it doesn't, the crisis will either be that it doesn't recognize anything at all and you won't be immune to anything, or the crisis will be that the body, the immune system starts to recognize self. And then we get one of this collection of autoimmune diseases that Dr. Banasra mentioned earlier on, juvenile diabetes or multiple sclerosis or myosthenia gravis or lupus or one of many, many diseases that some of you probably thought about it yourselves. So what I'd like to explain to you today is how the immune system does this, how it tells the difference between self and anything else. And I'd like to review some of the basic properties of the immune system so you can understand what I'm talking about and some of these will be points that I make will be reminiscent of what Dr. Banasra has already mentioned, but I'd just like to remind you about these points. So the first thing I need to do in order to explain how we tell the difference between ourselves and everything else is to explain to you how the immune system works when it's functioning normally and they're going to have a series of cartoons here to help you understand this. And what you need to know is that immunity is driven by lymphocytes.  And the critical thing about lymphocytes is that they have receptors on their surface, not drawn to scale here. This is my husband's picture. He took this picture. Actually there doesn't seem to be a little red dot going out. These lymphocytes each have receptors on their surface and we've only put one receptor on the surface but actually there's probably about 20,000 of these things on the surface of each lymphocyte. And what this diagram here illustrates is that for each lymphocyte the receptor is slightly different. It's slightly different in amino acid sequence. It's a protein. So each lymphocyte has about 20,000 of these receptors on its surface and they differ from one lymphocyte to another. And if you can imagine the magnitude of this you and I have about somewhere around a million million, 10 to the 12th or more of these lymphocytes in our bodies and each one of them more or less has a different receptor on its surface. So we're talking about a lot of different kinds of receptors here. Individually different from one cell to another. And as I expect probably many of you know when you get an infectious entering the body, for example this virus particle we've drawn here, by chance it happens to be able to bind to the receptors on some of these lymphocytes. In this case this lymphocyte right here, it fits into the amino acid sequence. It's able to bind to the amino acid sequence of the receptor on this lymphocyte. And the consequence of this is that this binding reaction right here between the virus and the receptor on the lymphocyte makes the lymphocyte divide and you end up with this clonal expansion of a lymphocyte which was able to recognize this foreign invading particle, the virus particle. And what happens is that at the end of the response we have many many more lymphocytes bearing receptors which can deal with this virus than you had when you started with. This can be an expansion of say a thousand fold in number. And that's in fact how you make an immune response to an antigen, a virus or whatever. And these cells at the other end here they're what's called effectal lymphocytes. They help you get rid of this virus particle in a number of ways that I don't have time to go into. Now this scheme which was for the immune response which was devised in the 50s by a number of people, MacFarlane Burnett, Neil Scioni and especially David Talmage who's in Denver right now developed this theory which is called the clonal selection theory is the basis for how we understand immune responses to occur. We have lots of lymphocytes. Each one has a different receptor on its surface and the antigen pulls out the ones that combine to and help deal with that particular antigen. Selects out the clonal cells. Now you can see that there's a problem with this system as far as self is concerned and that is that we believe that these lymphocytes and we know in fact generate receptors on their surface of random specificity. We don't know ahead of time what these receptors are going to be able to recognize and that means that you could generate lymphocytes bearing receptors that can recognize you and how do we deal with that problem given that the possible outcome of recognizing you is a whole collection of potentially damaging lymphocytes that can kill your cells or create inflammatory lymphocytes inflammatory hormone like molecules that could damage you. So why don't we generate lymphocytes that bear receptors that can recognize us and originally there were two hypotheses basically. One possibility is that for some reason or another we simply don't have the genes to code for the proteins to code for the receptors that could recognize bits of us. So we're just intrinsically unable our lymphocytes are just intrinsically unable to recognize us because we don't have the DNA to code for anti-self. That's one possibility. Another possibility is that somehow or another the immune system learns what self looks like and deals with the cells that could recognize self. So we know that we don't lack the genes we don't lack the DNA to code for proteins that can recognize us and the reason we know that is that my lymphocytes are very well able to reject tissue from my children actually they'd like to reject my children more often than not or my parents and of course my children's lymphocytes and my children are only too eager to reject me conversely. So and we know that we all share genes with each other so it would be almost impossible since I reject my lymphocytes would reject my close relatives tissues so effectively. It's almost impossible to imagine that genetically I lack the ability to recognize myself where I can certainly recognize my mother and my kids and so on very well. So this is sort of ruled out by that kind of reasoning the possibility that we lack that but there's actually another of course some very beautiful sets of experiments which tell us that the way we distinguish ourselves from everything else is because we learn to do that and the first experiment that was done to prove this was done by Ray Owen in 1945 which was the year I was born actually and what he showed was that if you have brothers and sisters calves who now we're talking about cows calves who are brothers and sisters normally they can reject skin grafts between each other very well but if they happen to have come out of the same pregnancy and they shared a placenta which that happens sometimes for calves they won't reject skin grafts one from the other the two twins don't reject skin grafts so somehow in utero the two calves learned that the other or at least they thought they learned that the other one was self too and their immune system won't reject the other calf anymore so we know that tolerance to self it's called tolerance the lack of ability to recognize self is something that your immune system learns now what I'd like to tell you about today is how the immune system learns what self is because it's actually a very simple and straightforward message as you'll see and quite clever the way biology has set this up but I have to tell you a little bit about the receptors on cells before I do that and the first thing I'd like to introduce is to remind you of something that Dr. Banathara told you earlier on today and that is that the immune system in fact actually has three different kinds of receptors three different kinds of lymphocytes and each lymphocyte bears an entirely different type of receptor for antigen on its surface so we have in us as human beings about a million million B cells and each B cell has on its surface a slightly different version of this molecule which is an immunoglobulin molecule that's an antibody molecule very similar to the kind of antibodies that you have floating around in your serum so B cells bear immunoglobulins and then we have these two different kinds of T cells this kind over here the gamma delta cell which I don't understand so I'm not going to talk about that and then this T cell here the alpha beta T cell which was the kind of T cell that Dr. Banathara was talking about and it's the cell that I'm going to discuss today as well included in this group of T cells are the CD4 bearing cells the ones that I think Dr. Gallo might discuss a little bit later on when he talks about AIDS infections because the target of the HIV of the immunodeficiency virus is a kind of this type of T cell an alpha beta bearing cell but what I'm going to say about alpha beta T cells applies to some extent or another also to B cells and I haven't any idea whether it applies to gamma delta cells so we won't worry about them now I need to tell you a little bit about the T cell receptor the alpha beta receptor and how it works the alpha beta receptor on T cells the question is how do you get all these millions and millions of different slightly different specificities out of these receptors how do you construct a receptor which on one cell has a particular amino acid sequence a particular structure and on another T cell is similar but not identical how's that done and it's the way it's done it's done genetically in rather a complicated way that I don't want to go into but basically it's done in a in a similar manner to a kind of a cheap Chinese restaurant and what that means is you know when you go to I don't know if this happens in the state so much but when you go to a cheap Chinese restaurant they say you should take choose one possibility out of this collection of A things like spare ribs or whatever and then one you can choose one thing out of column B etc etc and put together your dinner by combinations from different lists and that's basically what the T cell does it can choose for the alpha chain it can choose a V it's called V alpha out of a pool of about 50 or 60 different V alphas it can choose just one and then it can choose a J alpha and there are about 50 different J alphas and it can choose just one of the 50 and then there's this thing called N which I don't want to talk about too much but there's about 400 different choices for N so a T cell is choosing is making its alpha chain out of one of 50 and one of 50 and one of 400 and if you multiply that together the number of possible ways of putting an alpha chain together is very large so it has just like you could put your Chinese dinner together in a lot of different combinations the T cell does the same thing for its alpha chain and likewise for its beta chain it puts it together in a lot of different combinations and in fact this is a sort of ballpark figure it's plus or minus two orders of magnitude right they're probably around 10 to the 14th different ways that we could assemble a T cell receptor so that's why I say out of the million million T cells we have probably none of them have exactly the same receptor they've all made slightly different choices so I hope I made that clear the diversity of the receptor is put together by choosing one of many combination one of many possibilities at each of these positions so that's the first point I want to make about the T cell receptor that it's put together with these little segments each of which is different from one cell to another and then the other thing I need to tell you about the receptor and this is something that Dr. Banasra pointed out before is that the job here's the T cell up here see it's something huge right up here and the way the T cell receptor works is it sees little fragments of antigen little peptide fragments of antigen bound in the groove that Dr. Banasra shows you of the major histocompatibility complex molecule these are proteins that are on every surface the surface of many cells in the body and the way we've drawn this you can see here's the surface of the T cell receptor made up of all these segments that I said the T cell had a choice about and all of these segments contribute to the ability of the T cell up here to bind to this little antigenic fragment a little chunk of flu virus or polio virus or whatever it is bound in the groove right here so the point is that for most antigens flu or polio or chicken pox or tuberculosis or whatever ragweed um the T cell has to have exactly the right combination it has to have chosen the right V alpha and the right V beta and the right J beta and so on in order to be able to recognize that peptide so if you remember I told you there were lots and lots of different specificities what this means is that in an unprimed in animal in a normal individual who hasn't had this particular chicken pox virus before for example the frequency of responding T cells is very very low it's something like one in a million or one in a hundred thousand T cells is able to recognize that virus likewise for your own peptides for example for a peptide from your big toe or whatever it is we imagine that the frequency of T cells if you hadn't already been immunized with that big toe antigen would be very very rare one in ten to the fifth or so what that means is it's tremendously difficult to figure out what's happened to that T cell if you made a T cell that could recognize an antigen in your big toe how could we figure out what it was doing was it there was it present or not um is it in the body or is it just sort of sitting around ignoring your big toe because it's so rare in the population we can't tell which T cell to look at so the question is how do we devise a tool so that we can follow what happens to T cells that can recognize ourselves and uh a few years ago John Kappler and I happened on a tool it was a complete accident that turned out to be very useful and this was the special kind of antigen we discovered it's called a super antigen and you can see that regular bits of flu or TB or whatever antigen it is that's entered the body the foreign substance usually lie in this special little groove here but these super antigens that we discovered have a totally different way of interacting with the T cell receptor they bind onto these peculiar molecules these MHC molecules right here and they clamp the whole thing up the side by binding to a piece of the T cell receptor called V beta and they don't care about all these other variable bits they're only interested in this one little bit called V beta now in human beings we have about 50 different V beta genes 50 or 60 different V beta genes all together what that means is that about 2 percent of our T cells or more in fact can be recognized can recognize a super antigen because they have the right V beta on their surface the super antigen will grab them and so in man we can follow very easily the ability of a super the interaction of a super antigen with a T cell because it's easy to see what 2 percent of T cells are doing or at least it's tremendously easier to see what 2 percent of the cells are doing than one in a million cells is doing if you see what I mean and fortunately for us over the years people have made antibodies against these V's these different V betas and so we can actually follow T cells bearing particular V betas by seeing whether they react with antibodies that may seem a little complex but actually I have to tell you that technically it's much much easier to do than it was before and I didn't tell you this but in mice which is the animal we work with most of the time because we can't usually get human beings to volunteer to get immunized with these and you'll see why in a minute in mice there are only about 20 V betas so that means that 5 percent of the T cells in a mouse bear any given V beta or thereabouts and that's a piece of cake to see by comparison we're one in a million cells so you'll see the kind of experiment that we can do with this very useful reagents in a minute I thought I'd tell you just a few things about super antigens before we go any further the first thing is is that they have to bind to these MHC molecules which I don't want to go into in any depth right now and the second is just to remind you once again that the way these special antigens react with T cells is just through that special part of the receptor called V beta and now I need to tell you what super antigens are and some of them in fact everybody in this audience has met a super antigen I'll guarantee you because they are made by bacteria a whole collection of made by streptococcus and staphococci and they are the major cause of food poisoning in this country a staphococcal enterotoxin B for example was first worked on in Madison and sequenced in Madison it's a fairly medium-sized protein it's the kind of protein that when you eat an egg salad that's been sitting out too long makes you feel really bad for a short space of time for about a day it's that kind of food poisoning I'm talking about and we think in fact that the food poisoning has something to do with the fact that these toxins engage T cells and maybe I'll come back to that later on anyway so staphococci make about 11 or 12 different toxins which are super antigens one of them is the toxic shock toxin which gained some notoriety in the early 80s when relied tampons were brought in it was the cause of a considerable amount of toxic shock an epidemic of toxic shock in this country and the other point I want to make about these particular kind of toxins is that they have specificities for different V betas and in order to explain that I have to say that if you remember I told you that the T cell has a choice of one out of 20 V betas or whatever it is each V beta is different in sequence and the scientists who study them have numbered them so in mice we have V betas 1 through 19 and each number designates a slightly different amino acid sequence slightly different structure for that V beta so staphococcal enterotoxin B for example reacts with T cells in mice bearing V beta 7 and the members of the 8 family and in man if you were to eat staphentrotoxin B the T cells in you that would get all upset about this would be those bearing V beta 3 12 etc etc whereas in mice for example another staphococcal enterotoxin called SEA engages T cells bearing V beta mouse V beta 3 now it's not only bacteria that make super antigens there's also a collection of viruses that make super antigens and the most well-known of those are some viruses in mice which called which cause breast cancer they're called mouse mouse mammary tumor viruses MTV and one of the viruses that I'm going to mention later on is a virus that's passed through the mother's milk from the lactating mother to her baby in the milk and this virus carries with it a super antigen which engages T cells in mice bearing V beta 14 but these mouse mammary tumor viruses are especially interesting because over the course of the years some of these viruses have picked themselves up and put themselves into the DNA of the mouse themselves they've what's called integrated themselves so there are mice running around in your backyards which have integrated into their own DNA some of these mammary tumor viruses and they carry with them super antigens so as far as these mice in your backyards are concerned the virus in their DNA is them they can't tell the difference between the virus and themselves because they grow up with it it's in their DNA just like anything else in their DNA so we can use these viruses that are making super antigens that are actually in the mouse DNA itself to study what happens when a mouse T cell develops and sees a super antigen which is specific for that T cell now I'll try and illustrate that on the next slide first of all I want to tell you the people who did the work I'm going to talk about today as John Kapler and myself we run a lab together and our postdocs who were involved in this work were Yong-Won Choi and Terry Finkel and Lesha Gignardovitz and Jim McCormick and some of the work I'm going to talk about was in collaboration with Brian Kotzen if I have enough time and another postdoc who mysteriously didn't show up on this slide whose name is Xavier Palliano so how do we tell the difference between ourselves and everything else given that we could make lymphocytes that could recognize us so basically over the years there are three theories that have come forward to account for tolerance to self and they are one possibility is that T cells while they're developing go through a stage where if their receptors get engaged by something that cell will die and since self is always present when T cells are developing self is always present all T cells while during development will go through this screen of whether or not they can react with self and if the clonal elimination theory the clonal deletion occurs they will die during that developmental stage I don't know if I said that clearly enough but I'll say it again later on another possibility is that contact with self during this sensitive developmental stage will cause the cell not to die but simply to become inactivated so it'll still sit around but it won't actually die but it won't be able to do anything threatening to the host anymore because it's inactivated by this contact during an early developmental stage and another possibility that's been suggested is that somehow or another potentially auto reactive T cells are kept at bay by other cells that keep them under control called suppressor cells so what I'd first like to show you is some evidence that at least for many self antigens this first mechanism clonal deletion is a very important one in maintaining self tolerance in order to do this I first have to explain to you how T cells develop T cells are formed in the thymus this gland right here in your chest and they go through the cells enter the thymus from the bone marrow or the yolk sac or fetal liver and at that point they don't have any receptors on their surface at all and when they get to the thymus the environment of the thymus induces the appearance of receptors each different on one T cell from another on the surface of these cells and then they go through three or four days of stage here which doesn't have very much receptor on its surface it's called an immature thymocyte and then the cells go through a process that Dr. Bonaster I've alluded to called positive selection that I won't discuss and they turn into mature lymphocytes first in the thymus medulla right here and later on in the periphery in the spleem and lymph nodes and peripheral blood of the animal so the T cell develops in the thymus and it goes through an immature phase in the thymus so now this is a complicated slide to explain what we have on this axis is the percent what we can do is look in individual mice and ask how many cells do they have how many T cells do they have which have on their surfaces for example this V beta V beta 6 and in a normal animal it's around about 12 percent of all the T cells bear V beta 6 on their surface now some mice are infected not are infected have in their DNA integrated into their DNA one of these mammary tumor viruses I told you about that makes a super antigen which can react with T cells bearing V beta 6 so that the T cells in that animal are growing up in the constant environment of this mammary tumor virus making a super antigen specific for V beta 6 and we can ask what percentage of the T cells in that animal bear V beta 6 and the answer is it's much less than the expected percentage instead of 12 percent which is what we've called 100 percent right here it's very confusing picture instead of 12 percent is less than 1 percent and this is we're looking at the periphery of the animal now in the lymph nodes and spleen so that the expression of this super antigen that as far as the mouse is concerned is self in the animal has caused the disappearance of all the T cells or most of the T cells bearing V beta 6 in that animal now when did that happen did it happen when the T cells got to the periphery or did it happen in the thymus and this is a slide that shows you that it happened in the thymus if you could just follow along here this is the mice these are mice which have that MTV that encodes a V beta 6 reactive super antigen you can see they have a very low percentage of the mature thymocytes bearing this antigen bearing V beta 6 and they have a reduced less than expected number of the immature thymocytes bearing this V beta so the summary of this kind of experiment is that thymocytes do indeed go through a stage right here while they're immature about halfway through their immature phase the point of this red line right here in which they become sensitive to deletion and if their receptors at that time become engaged with something that can react with those receptors that is a signal for death of those thymocytes and therefore those particular cells because they're dead can no longer proceed through to the mature thymocytes stage and to the periphery and we see this is a deletion of that particular subset of cells so in this particular case self has caused clonal deletion of potentially reactive from T cells in the thymus so just to summarize what I've told you so far and to make sure to really ran this point through what this experiments have shown is that during their development T cells go through a stage when if their receptors bind antigen bind something those cells will die because our self is always there self is always present all developing cells as they go through this kind of adolescent rebellious stage will get checked out for their ability to be killed by antigen by self and if they their receptors react with self those cells will die and the critical point is that infectious things from the outside like for example polio virus are not always present in the body the child is born without not infected with polio virus and therefore during the early stages of that child's life its thymus will produce T cells that can react with polio virus antigens later on when the child becomes confronted with polio vaccine I hope or polio virus perhaps those T cells that are now already mature can help respond to and get rid of the virus as I showed you on a much earlier slide at the same time of course the T cells which are developing right at that very moment now I think the polio virus is self because it's in the body at the very time that those particular cells are developing and they'll die but it doesn't matter because by that time the child has already accumulated enough T cells mature T cells to help get rid of the virus so that's the theory to account for the data that I've just shown for you that I've just shown you about clonal deletion and I have sort of some sort of hysterical slides right now it is possible that not all antigens get into the thymus for example you could imagine that things that were in your big toe how would they ever get to the thymus well some of them do get carried back to the thymus on the surface of macrophages or whatever or B cells or something and and maybe able to tolerize T cells there but perhaps there are some kinds of antigens that don't get carried back very well for the thymus and how do we keep our lymphocytes tolerant to those and especially critical in this regard are various antigens that are in your brain or perhaps in your joints and I should tell you at this point that it's not clear that we are in fact tolerant to the antigens that are in our brain but I'd like to deal with a few more antigens that are elsewhere in the body first and so because we now have these tools the super antigens to help us in our experiments I'd like to illustrate this question of peripheral tolerance how we become tolerant to antigens outside the thymus with a couple of experiments here now in this case we've taken a mouse and injected an adult mouse and injected it with one of those bacterial super antigens this is SeA which if you remember in mice stimulates T cells bearing Vbeta3 so we injected the mouse with SeA right here on day zero and then we asked what percentage of T cells in the lymph node and spleen have on their surface this Vbeta3 with which SeA can react and what you see is there's a big spike as the lymphocytes proliferate in response to this antigen in number and then they fall off in fact they a fair number of them die and disappear and we're left with some cells left over at the end here which is somewhat less than we started with so an acute confrontation with antigen goes almost pretty much as you would have predicted you see a response as the lymphocytes divide and then they die away but if we put the antigen in chronically that is sort of every other day pretending as though it was an antigen in your big toe and you kept on challenging lymphocytes with it what we see is that if we put in a lot of it there's a spike of response but if we put in just a little antigen the lymphocytes just disappear the Vbeta3 bearing cells just disappear and eventually dwindle away to nothing so there's some kind of mechanism whereby a chronic antigen which is what self is causes the disappearance of target cells in the periphery and I have to tell you now we have no idea what this mechanism is and we certainly would like to know what it is because you could imagine the therapeutic value of understanding this mechanism in autoimmune diseases if we knew what the antigen was and we knew how to get this chronic disappearance business to work we perhaps could deal with autoimmunity much more effectively than we do right now so the immune system once again is dealing with potentially autoimmune C cells on the basis that the antigen is being continuously administered to them and as I said we don't understand what this phenomenon is but it looks like something real if and we wish that we understood it better than we do I'd like to point out one other thing about this mechanism and this is a little difficult to grasp from this slide so I'll just tell you and that is that we can interfere with that disappearance in response to peripheral antigens if we put in an inflammatory agent at the same time so we're chronically giving this mouse staphocontrococcal entrotoxin A for example and making the T cells bearing VB to 3 disappear and if at the same time we give the mouse some kind of really strong stimulus an inflammatory stimulus in this case an adjuvant called complete Freund's adjuvant they don't disappear after all so maybe that's another phenomenon that has something bearing on autoimmune diseases and how they stop so to summarize what I've told you so far acute exposure to super antigens of peripheral mature T cells causes them to expand and then disappear a little bit and if we expose mature T cells chronically to a super antigen and presumably a regular antigen too this causes them to die and in an inflammatory process such as complete Freund's adjuvant an inflammatory agent can interfere with this deletion and we are thinking along the lines that some phenomenon like this or related to this may have some bearing on the induction of autoimmunity in patients now I don't think I have time to show you this so I'll skip it and just come to the summary and that is that what I didn't have time to show you is that under some special circumstances instead of getting deleted T cells can get inactivated so basically what I've described so far is this mechanism of clonal deletion adenolgium in the thymus or in the periphery self can cause the disappearance of target cells and also under certain circumstances um self can instead of courtesing these target cells to disappear can just make them inactive and what I haven't discussed with you is this third mechanism of suppression and I'm not going to do that today what I wanted to finish up with very briefly is some description of how this is relevant to disease in man and what I'm going to show you is very hyper their real data but the explanation of the data is pretty hypothetical as you will see and a little complicated so I hope you'll follow along with the results now if you remember in man I told you there are about 60 different V betas and each T cell bears a different V beta in this assay that I'm showing here we only actually measured 20 of them or families of 20 so what's shown on this slide is the percentage of T cells in two individuals that bear different V betas on their surface so for example you can see that both of these individuals have a fair number of T cells bearing V beta 13.1 which is just a name for an amino acid sequence on their surface so this is a commonly used V beta by these two people and by most of the people in this audience um and I just wanted to point out a couple of things of this about this of general interest the V betas are not sort of equally often used some of them are used very frequently like this 13.1 by T cells and some of them are hardly ever used like we none of us have any T cells hardly at all bearing V beta 10 and I don't know why it should be such a poor despised relative but it just isn't often used the other point I wanted to make is that these two individuals are not identical for example the dark one the dark person right here has hardly any T cells bearing V beta 3 and the hatch person here who's me who's infinitely superior has many more T cells bearing V beta 3 um in her circulation and we don't know what that what caused that we don't know the significance of that it's just a fact right now uh and the scientific and medical consequences of that we don't understand now what I wanted to show you here is a patient this is some experiments that the Xavier Pagliata postdoc in our lab and Brian Kotzen published about a year ago to dealing with the T cell repertoire in people with rheumatoid arthritis which is an autoimmune disease of the joints as I expect some of you know and he Xavier was trying to figure out whether there was anything peculiar about T cell receptors about the T cells in patients with rheumatoid arthritis so what he did was he collected the T cells from the peripheral blood and the synovial fluids that is the rheumatic joints of patients uh of the same patient and compared what kind of T cell was in the peripheral blood to the kind of T cell that was in the synovial fluid and in the dark bars the dark bars show the percentage of T cells bearing different V betas um in the peripheral blood again he was looking at V betas 1 through 20 and the hatched bars show the same V betas measured in the synovial fluid the percentage of T cells bearing them and what you can see is that on the whole things are rather similar except V beta 14 um where there's definitely a lot more T cells in the synovial fluid than there are in the peripheral blood bearing this V beta and this wasn't true simply for this one patient it was true for every patient we looked at this is a collection of room patients with rheumatoid arthritis right here the dark bars in every case of their peripheral blood and the hatched bars is what's happening um in their synovial fluid and you can see for all of these there are more V beta 14s in the synovial fluid than in the peripheral blood here are some control patients who had a different kind of arthritis in their joints not rheumatoid and here are a collection of individuals who didn't have arthritis and we can't look at the T cells in their synovial fluid because first of all it's not ethical to go sucking things out of people's knees if they don't have something wrong with them and secondly they don't have any if you're a healthy individual you don't have any T cells in your knees anyway so that was a good reason not to do it apart from the ethical considerations right there but actually the point I really want you to notice from this slide is not the difference between the synovial fluid and the peripheral blood but the fact that some of these patients with arthritis had no detectable T cells bearing V beta 14 in their peripheral blood at all it was like they were all gone and that wasn't true for any of the normal controls that we looked at so there was something about this disease that was removing the T cells bearing V beta 14 um from the blood um and one explanation for this is that um there's a super antigen specific for V beta 14 sitting in the joints of these patients that's sort of sucking all the VB to 14 into the joints all the VB to 14 pairings T cells into the joints so there's nothing in the peripheral blood they're all in the joints well let me tell you there isn't enough room in your knees for all the VB to 14 bearing cells you have in your body simply wouldn't fit so that's probably not the explanation and in fact I don't have time to go into it we we know that that's not true it's not that there's a super antigen in the joints attracting all the T cells in there what the disease looks much more like what the phenomenon looks much more like is the expression of some sort of chronic super antigen in these patients bodies which has caused deletion remember I told you that chronic expression of antigen causes deletion of peripheral T cells so it's called all the VB to 14 cells in the periphery of these patients to disappear and the only ones that are left behind are a few that could be rescued because they're sort of hidden in the joints and responding to antigen in the joints so I've written the theory down on this slide I think here we go this is such an exciting way to show this slide what this slide is going to show you is the hypothesis we developed to account for the fact that some arthritis patients don't have any T cells bearing this particular VB in their joints and theory goes the theory goes like this that these people are wandering around fairly normal they have a few T cells in them already that have snuck through all these processes of tolerance that I told you about so they just didn't see joint antigen in their thymus and the joints are all nicely encapsulated so they never see the antigens the collagen or whatever it is that's in the joints in their periphery so a few T cells go through the thymus they mature and they come out and potentially these T cells could react to that individual's collagen antigens for example joint antigens if they ever saw them but because they're never exposed to the joint they never do anything much and they're not activated they're not excited about anything they're not doing anything then one day this individual catches something that chronically expresses a super antigen specific for VB to 14 and this causes the activation of all the VB to 14 cells in this individual and one of the things that happens when T cells become activated is they become much more mobile and they can travel around better so the theory goes that these T cells can sort of some of the mobile ones go through the joints and the VB to 14 cells that could recognize joint antigens stay there and initiate the process of disease the process of rheumatoid arthritis in the joints meanwhile the chronic expression of this VB to 14 specific super antigen deletes all the T cells everywhere else in the body because it's there for such a long time and they the T cells begin to think that the super antigen is self if we could skip that slide and just do I get extra minutes for this so here's the theory I don't want to go through it again you you heard it the these people who are rheumatoid arthritis is an MHC linked disease in in this country it's linked to DR expression of an MHC antigen MHC protein called DR4 all those individuals I showed you in fact were positive for the DR4 MHC type so what we're saying is in some people who are DR4 positive these T cells sneak through tolerance somehow or another and they sit around and don't do any damage all of a sudden the individual becomes invaded by an exogenous antigen a super antigen that's specific for VB to 14 it activates the cells the joint specific VB to 14 bearing cells whiz off to the joints and start to cause the disease rheumatoid arthritis meanwhile in the periphery the rest of the cells become deleted and then other T cells join the VB to 14 bearing cells in the joint and you eventually develop the symptoms of rheumatoid arthritis this is very hypothetical and almost certainly not right but it's a sort of idea on which I don't mean to belittle it but what I mean is that you're never right the first time but you sort of approach rightness gradually so this is a sort of a a hint about how these diseases might start it's not the absolute answer but it's the kind of thing that makes you feel as though you have your foot in the door and you can start prizing the door open a bit more and maybe see more glimmerings of the truth as you go further inside but what I wanted to finish off with is the last slide which describes how the immune system tells the difference between self and everything else because this is the absolute critical feature that I wanted you to take home with yourself with you and that is we tell ourselves from everything else because we are always there and the infectious agents are only there some of the time and that is the critical element which defines self from non-self thank you very much


we have received questions online and also here in the audience so I'll read this one from the audience inshallah as someone who wants to be a parent one day inshallah how do you gently teach your kids about Islam especially in today's world there are a lot of bad influences that can take us away from the deen anybody want to take it yeah so as far as the bad forces that can take people away from the dean something that I'm with that he was a very respected scholar uncle in the community here one of my friends asked him you know what do we do with all the horrific horrible things that we see in the world that are going on around us that cause us to be afraid causes to worry for the sake of our children what do we do about that and he said something really interesting he said in every storm that alas Panathala has created he's created an eye the eye of the storm right where everything's calm everything's safe nothing's flying around to pray that Allah keeps you in the eye of the storm so it's true there are things going on all around us that are horrific there's also a lot of beauty all around us as well inshallah and we pray to Allah every day and our children see is praying for this as well that Allah keeps us in his protection keeps us in the eye of the storm I mean as far as how to gently teach the children about the dean the children have to see a lot of joy in the practice of the dean and they have to see that Islam works they have to see that Islam gives you a dignified life and it gives you a clean life and it gives you a better option for a way of living than what others may be choosing all around them and eventually they will come to see the difference between right and wrong because what's right has been presented in such a beautiful way to them their entire lives so I had somebody once say well I don't want to say anything that sounds like well okay I had somebody once say to me that you're one of the only happy Muslims I know so you know and they were saying that's why they wanted me to speak at a public event so they were like you're a happy Muslim you're a happy woman who's a Muslim and I thought that was a very sad statement to make because why don't we look happy why I know that we're worried about the next life and we're worried about how to get through this world and safety Jalla but our Dean gives us so much beauty and so much dignity and so much grace and so much hope there's so much hope and there's so much beauty around us so instead of just focusing on all the negative that's out there also get our kids to see all the positive right so if kids see that Islam is the reason the parents treat each other with respect and Islam is the reason we smell nice and Islam is the reason our homes are clean and Islam is the reason we treat our elders with respect and Islam is the reason we pay our bills on time Islam is the reason we keep our promises they will eventually choose it for themselves inshallah we want them to choose it because they recognize that it's the you know Huck obviously but when they also see that it gives you a dignified life they will choose it for themselves inshallah and teach it with a smile


kufingsa kuwa kufa felekua na uluza kuadi hisha tia nama integration kufa wa dalia wakupo hini kutu afid Holo kufa l tale kufa kufa kufuswa kila kufa kama kufuswa wakumu naka kufuna kufun mbo kufusa hatiwa kufuswa kufua hisha kufa kufa But I can confidently confirm that I was with Comrade Tulifeta Adi just behind me here when he was being taken to the four scotters by three police officers where we have been informed that he will be interrogated and asked questions concerning the very case upon which he was arrested. So I thought it will be prudent that we keep you posted because every Zambian lives matter and together as Zambia we have a duty to make sure that we make Zambia great. This is Kasun Demwenda, the President for the Economic Freedom Fighters. Whatever we want to go our feet will definitely take us there. Let us continue supporting the cause for democracy and let us make sure that the law is not abhazardly misused. He has already spent a night in the police cells and we have got enough laws in this country that should not allow him to spend more days behind these cells without him being charged. And we remember President Akande Chidema was very instrumental in condemning such kind of arrests which do not culminate into a quick charge. Right now Comrade Tulifeta Adi has not been charged. He is being held and within a few hours by this evening the grace period by which the law allows the police to hold somebody in their cells without being charged will be elapsing. So we should make sure that we follow this government and its actions closely that they do not take us back to the very things that they condemn strongly and heavily when they are in the opposition. That they should not use the law that they condemn to be bad now to be using it against the citizens. Of course that does not mean then as citizens and as political leaders should abuse the law. We have a duty to uphold the law and in this case we should make sure that we do not only support things that against the state but even things that the state perpetuates against its citizens should equally be condemned. So I have come here and I can assure you will keep you updated to keep you posted even in this case where Comrade Tulifeta Adi complained that there was an invasion at his house and he strongly suspected that the people that invaded his privacy and his house with ammunition, with guns were from the army. So we follow this issue closely and we make sure that the government does not harm him but lets the law take its course. Thank you, God bless you and God bless Zambia. Thank you.