Explanation Given information: The side of the square piece of cardboard is 10 inches. A square is cut from each corner of cardboard and sides are bent to form a box with open top. Calculation: Assume x (in inches) be the length of each square cut from the corners of cardboard. A square is cut from each corner of the cardboard and a box is formed by folding up the remaining length. The length of the box is l = 10 − 2 x inches. The width of the box is b = 10 − 2 x inches. The height of the box is h = x inches. The volume of the box is, V = l × b × h = ( 10 − 2 x ) × ( 10 − 2 x ) × x = 100 x − 20 x 2 − 20 x 2 + 4 x 3 = 4 x 3 − 40 x 2 + 100 x The derivative of V with respect to x is, d V d x = d d x ( 4 x 3 − 40 x 2 + 100 x ) = d d x ( 4 x 3 ) − d d x ( 40 x 2 ) + d d x ( 100 x ) = 4 d d x x 3 − 40 d d x x 2 + 100 d d x x = 4 ( 3 x 3 − 1 ) − 40 ( 2 x 2 − 1 ) + 100 ( 1 x 1 − 1 ) Further simplify the above equation, d V d x = 12 x 2 − 80 x + 100 Substitute 0 for d V d x to find the all critical points, 12 x 2 − 80 x + 100 = 0 3 x 2 − 20 x + 25 = 0 3 x 2 − 15 x − 5 x + 25 = 0 3 x ( x − 5 ) − 5 ( x − 5 ) = 0 Simplify the above equation to find the value of x , 3 x ( x − 5 ) − 5 ( x − 5 ) = 0 ( x − 5 ) ( 3 x − 5 ) = 0 Apply zero factor property to find the values of x . Substitute 0 for ( x − 5 ) , ( x − 5 ) = 0 x = 5 Substitute 0 for ( 3 x − 5 ) , ( 3 x − 5 ) = 0 x = 5 3 Note the value of x = 5 will make the value of l and b equal to 0

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Chapter 4, Problem 54RE

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The maximum volume of the box formed from a square piece of cardboard.

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