Explanation Given: The dimension of the cardboard is 3 foot by 8 foot . Formula used: Volume function of the resulting open box will be: V ( x ) = l × b × h Here, l , b , h refers to the length, breadth, depth of the open box, respectively. And, Critical point theorem, if a function f ( x ) is continuous on a given interval ( a , b ) and has only one critical number in the interval ( a , b ) , say, x = c . 1. If f ( x ) has a local maxima at x = c , then it will be the absolute maxima of f ( x ) on the interval ( a , b ) . 2. If f ( x ) has a local minima at x = c , then it will be the absolute minima of f ( x ) on the interval ( a , b ) . Calculation: The figure for better understanding of the question is given below: (a) (b). Write the expression for dimension of the resulting box (fig (b)): l = ( 8 − 2 x ) b = ( 3 − 2 x ) h = x Therefore, volume function of the open box will be: V ( x ) = ( 8 − 2 x ) ( 3 − 2 x ) x Simplify the above expression: V ( x ) = ( 24 − 22 x + 4 x 2 ) x = 24 x − 22 x 2 + 4 x 3 Since, x is the side of square, it is positive. Therefore, the domain of the volume function is the open interval ( 0 , ∞ ) . Determine the critical number for the volume function. Differentiate the above function: V ′ ( x ) = 24 − 22 ( 2 x ) + 4 ( 3 x 2 ) = 24 − 44 x + 12 x 2 To find the critical number, put V ′ ( x ) = 0

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ISBN: 8220106844557

Author: MULLINS

Publisher: PEARSON

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Chapter 12.3, Problem 39E

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To calculate: The size of square that should be cut from each corner for maximum volume.

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Chapter 12.3, Problem 38E

Chapter 12.3, Problem 40E

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EBK MATHEMATICS WITH APPLICATIONS IN TH

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