Solution Quiz 8

# Solution Quiz 8 - x 9. Now, we solve the problem precisely....

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Version A Solutions 1. A square sheet of paper 18" on a side is made into an open box (i.e. there's no top), by cutting squares of equal size out of each corner and folding up the sides. Find the dimensions of the box with the maximum volume. Length = ____ ", width = ____ ", height = ____ ". Solution: The volume of a rectangular parallelepiped (a box) is given by V = l w h . The height is h = x , while the length and width are l = w = 18 − 2 x . Thus, we can write the volume in terms of the one variable x as V = V ( x ) = (18 − 2 x ) 2 ( x ) = 4 x (9 − x ) 2 . Notice that since x is a distance, we have x 0. Further, we have x 9, since cutting squares of side 9 out of each corner will cut up the entire sheet of paper. Thus, we are faced with finding the absolute maximum of the continuous function V ( x ) = 4 x (9 − x ) 2 on the closed interval 0

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Unformatted text preview: x 9. Now, we solve the problem precisely. We have V '( x ) = 4(9 x ) 2 + 4 x (2)(9 x )(1) Product rule and chain rule. = 4(9 x )[(9 x ) 2 x ] Factor out 4(9 x ). = 4(9 x )(9 3 x ). So, V has two critical numbers: 9 and 3 and these are both in the interval under consideration. We now need only to compare the value of the function at the endpoints and the critical numbers. We have V (0) = 0, V (9) = 0, and V (3) = 432. So, the maximum possible volume is 432 cubic inches. We can achieve this volume if we cut squares of side 3 out of each corner. Observe that the dimensions of this optimal box are 12.0&amp;quot; long by 12.0&amp;quot; wide by 3&amp;quot; deep. Correct answer is: Length = 12.0&amp;quot;, width = 12.0&amp;quot;, height = 3&amp;quot;....
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## This note was uploaded on 05/06/2008 for the course MATH 201 taught by Professor Chuckrow during the Spring '08 term at CUNY City.

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Solution Quiz 8 - x 9. Now, we solve the problem precisely....

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