Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.7 - Optimization Problems - 4.7 Exercises - Page 337: 16


The minimum possible cost of materials is $~~\$163.54$

Work Step by Step

Let $x$ be the width of the base, let $2x$ be the length of the base, and let $h$ be the height of the container. Note that $x \geq 0$ and $h \geq 0$ We can express $h$ in terms of the volume $V$ and $x$: $V = 2x^2h$ $h = \frac{V}{2x^2}$ We can write an expression for the cost: $C = 10(2x^2) + (6)(2)(2xh)+ (6)(2)(xh)$ $C = 20x^2 + 36xh$ $C = 20x^2 + (36x)(\frac{V}{2x^2})$ $C = 20x^2 +\frac{18V}{x}$ We can find the point where $C'(x) = 0$: $C(x) = 20x^2 +\frac{18V}{x}$ $C'(x) = 40x -\frac{18V}{x^2} = 0$ $40x =\frac{18V}{x^2}$ $40x^3 =18V$ $x^3 =\frac{18V}{40}$ $x =\sqrt[3] {\frac{18V}{40}}$ $x =\sqrt[3] {\frac{(18)(10~m^3)}{40}}$ $x =\sqrt[3] {\frac{9~m^3}{2}}$ $x =\sqrt[3] {\frac{9}{2}}~m$ $x = 1.65~m$ When $0 \lt x \lt 1.65$, then $C'(x) \lt 0$ When $x \gt 1.65$, then $C'(x) \gt 0$ Thus, $x=1.65~m$ is the point where $C$ is a minimum. We can find the minimum possible cost: $C(1.65) = 20(1.65)^2 +\frac{(18)(10)}{1.65}$ $C(1.65) = 163.54$ The minimum possible cost of materials is $~~\$163.54$
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