CABChp4-4AampB4NTs - CalcAB Chp 4-4A Optimization Models...


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CalcAB Chp 4-4A Optimization Models Notes Students, on your own papers you will be writing the actual calculation work and answers to each problem. The notes here are intended to guide how/why to address different kinds of problems. 1. OPTIMIZATION problems ask you to solve for a measurement or quantity that will MAXIMIZE an amount such as volume, area, or profit. Or, MINIMIZE an amount such as cost, distance, or perimeter. The problems will mention (or imply) something is to be "maximized" or "minimized". We will be working with extreme value points ; the maximum and minimum turning points studied in Chp 4-1 and Chp 4-3. The x-coordinates of extreme value points are called critical numbers. They are calculated by setting a first derivative equal to 0 . (Application problems are real world type of situations; it is not appropriate to consider the derivative as undefined.) 2. Ex(volume of an open top box)— An open top box is to be made from a square piece of cardboard, 8 inches on a side, by cutting equal squares from each corner and turning up the sides (see figure). (a) Write the volume as a function of x. (b) What dimensions of the box will hold maximum volume? (c) What is the maximum volume? (a) Equation for volume Need the geometry formula for volume of a rectangular prism. V = lwh Write algebraic expressions to use for the sides of the prism . To turn a rectangular piece of cardboard into a box the four square corners need to be cut out; all with the same size side x. Then, fold up the four flaps to create an open-top box (rectangular prism). The length of the original cardboard is 8 inches when the squares are cut out from each end. That means 2x is cut off each side of the original cardboard. Need algebraic expressions for the length, width, and height of the prism formed. (b) Dimensions of the box with largest volume. Unit of measure for length is inches . Need to solve for x ; x is the square cut out of the original 8-inch cardboard. This problem is only interested in the box with the largest volume. In Part(a) the volume equation developed represents the volumes of all possible boxes. Look at the graph of V(x). It has two turning points. (You will need to change the window setting to show y-values up to 45.) The turning point that is a maximum is the point needed . The x-coordinate can be used to get the dimensions of the largest box and the y-coordinate is the maximum volume .
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