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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&quot; long by 12.0&quot; wide by 3&quot; deep. Correct answer is: Length = 12.0&quot;, width = 12.0&quot;, height = 3&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.
 Spring '08
 CHUCKROW

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