Explanation Given Information: The volume of Drum tight containers is 62 .5in 3 . Formula Used: Formula used: The concept of Max-Min Principle-2 is taken into consideration which is as follows: Suppose f is a function such that f ' ( x ) exists for every x in an interval l and there is exactly one critical value c in l , for which f ' ( c ) = 0 Then, f ( c ) is absolute maximum value over l if f ' ' ( c ) < 0 Or f ( c ) is absolute minimum value over l if f ' ' ( c ) > 0 Proof: Let x represent the length and width and y represent the height of the drum. Consider the diagram below, Find the surface area of the rectangular box by adding the area of the base and the four sides, S = x 2 + 4 x y The volume must be 62 .5in 3 . V = l ⋅ w ⋅ h = x 2 y = 62.5 To express S in terms of one variable, solve x 2 y = 62.5 for y, y = 62.5 x 2 Substitute y in surface area equation, S ( x ) = x 2 + 4 x ( 62.5 x 2 ) = x 2 + 250 x = x 2 + 250 x − 1 Since surface area is defined only for positive numbers, therefore, Interval restricts to ( 0 , ∞ ) . Find the first derivative of surface area function, S ' ( x ) = 2 x − 250 x − 2 = 2 x − 250 x 2 This derivative exists for all x in the interval ( 0 , ∞ ) . So, set this derivative equal to zero, S ' ( x ) = 0 2 x − 250 x 2 = 0 x 3 = 125 x = 5 Hence the critical value obtained is x = 5 which lies in the interval ( 0 , ∞ )

Calculus and Its Applications, Books a la Carte Plus MyLab Math Access Card Package (11th Edition)

11th Edition

ISBN: 9780133862386

Author: Marvin L. Bittinger, David J. Ellenbogen, Scott J. Surgent

Publisher: PEARSON

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Chapter 2.5, Problem 75E

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To prove: The dimensions and surface area of an open-top, square-based, rectangular box having volume of 62.5in3 such that the surface area is minimum.

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Chapter 2.5, Problem 74E

Chapter 2.5, Problem 76E

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To prove: The dimensions and surface area of an open-top, square-based, rectangular box having volume of 62 .5in 3 such that the surface area is minimum.

Solution Summary: The author explains the concept of Max-Min Principle-2, which is taken into consideration.

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To prove: The dimensions and surface area of an open-top, square-based, rectangular box having volume of 62.5in3 such that the surface area is minimum.

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