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Applied Problems in Maxima and Minima

Date: 08/30/97 at 23:40:40
From: Anonymous
Subject: Applied problems in maxima and minima

tangent together with the coordinate axes determine a triangle of 
minimum area.

Answer: (2 (3)^(1/2) / 3, 8 / 3)

Please show me how to get the answer.

Date: 08/31/97 at 12:38:45
From: Doctor Anthony
Subject: Re: Applied problems in maxima and minima

We can simplify the working a little by considering the curve y = x^2, 
and consider the triangle of minimum area formed by a tangent to the 
curve, the y axis, and the line y = 4.  This allows us to use very 
simple parametric coordinates (t,t^2) to represent the curve.

  dy/dx = (dy/dt)/(dx/dt) = 2t/1  =  2t

The equation of the tangent is  

                    y-t^2 = 2t(x-t)

This meets the y axis where x = 0, so 

                    y-t^2 = -2t^2

                        y = -t^2

as expected this is a distance t^2 below the x axis. The vertical side 
of the required triangle will therefore be of length 4+t^2.

The tangent meets y = 4 where  4-t^2 = 2tx-2t^2
                             4 + t^2 = 2tx

                                   x = (4+t^2)/2t

Area of triangle  = (1/2)(4+t^2)(4+t^2)/2t

                A = (1/4t)(16 + 8t^2 + t^4)

                  = (1/4)(16/t + 8t + t^3)

            dA/dt = (1/4)(-16/t^2 + 8 + 3t^2)  = 0 for max or min.

So       3t^2 + 8 = 16/t^2

 3t^4 + 8t^2 - 16 = 0

  (3t^2-4)(t^2+4) = 0    and so  t^2 = 4/3  

                                   t = 2/sqrt(3)

Thus the x coordinate is 2/sqrt(3).  This is the same x value as 
for the equation y = 4 - x^2, and therefore we can substitute 
x = 2/sqrt(3) in this equation to find the y coordinate.

  y = 4 - 4/3 = 8/3

Coordinates of required point   [2/sqrt(3), 8/3]
-Doctor Anthony,  The Math Forum
Associated Topics:
High School Calculus

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