To determine The maximum area that can be enclosed with 800 ft of fencing. Explanation Let x is the side of rectangular storage and y is the length of rectangular storage. Perimeter of the rectangular storage is P = 2 x + y . Substituting, P = 800 in P = 2 x + y : P = 2 x + y 800 = 2 x + y y = 800 − 2 x The area of rectangular storage A = x ⋅ y Substitute, y = 800 − 2 x in A = x ⋅ y : A = x ⋅ y = x ( 800 − 2 x ) Differentiate A with respect to x : A = x ( 800 − 2 x ) d A d x = ( 800 − 2 x ) + x ( − 2 ) = 800 − 4 x For maxima and minima d A d x = 0 . d A d x = 0 800 − 4 x = 0 4 x = 800 x = 800 4 x = 200 Differentiate d A d x with respect to x

Basic Technical Mathematics with Calculus (11th Edition)

11th Edition

ISBN: 9780134437736

Author: Allyn J. Washington, Richard Evans

Publisher: PEARSON

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Chapter 24.7, Problem 20E

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The maximum area that can be enclosed with 800 ft of fencing.

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Chapter 24.7, Problem 19E

Chapter 24.7, Problem 21E

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Chapter 24 Solutions

Basic Technical Mathematics with Calculus (11th Edition)

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