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: 17

Answer

The minimum possible cost of materials is $~~\$191.28$

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: $C = 10(2x^2) + (6)(2)(2xh)+ (6)(2)(xh)+(6)(2x^2)$ $C = 32x^2 + 36xh$ $C = 32x^2 + (36x)(\frac{V}{2x^2})$ $C = 32x^2 +\frac{18V}{x}$ We can find the point where $C'(x) = 0$: $C(x) = 32x^2 +\frac{18V}{x}$ $C'(x) = 64x -\frac{18V}{x^2} = 0$ $64x =\frac{18V}{x^2}$ $64x^3 =18V$ $x^3 =\frac{18V}{64}$ $x =\sqrt[3] {\frac{18V}{64}}$ $x =\sqrt[3] {\frac{(18)(10~m^3)}{64}}$ $x =\sqrt[3] {\frac{45~m^3}{16}}$ $x =\sqrt[3] {\frac{45}{16}}~m$ $x = 1.41~m$ When $0 \lt x \lt 1.41$, then $C'(x) \lt 0$ When $x \gt 1.41$, then $C'(x) \gt 0$ Thus, $x=1.41~m$ is the point where $C$ is a minimum. We can find the minimum possible cost: $C(1.41) = 32(1.41)^2 +\frac{(18)(10)}{1.41}$ $C(1.41) = 191.28$ The minimum possible cost of materials is $~~\$191.28$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.