Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.1 Four Ways to Represent a Function - 1.1 Exercises - Page 22: 61

Answer

$S(w)=w^2+\frac{8}{w}$ $w>0$

Work Step by Step

Formulas used volume $v=lwh$ area $a=lw$ total surface area = base area + 4(side area) We are told the volume of the box is $2m^3$, so $v=2$. We are also told the base is square. So we know the $4$ sides of the box will have equal areas. Fill in the volume formula with what we know $2=w*w*h$ Multiply $2=w^2h$ Divide $\frac{2}{w^{2}}=\frac{w^{2}h}{w^{2}}$ Simplify $\frac{2}{w^{2}}=h$ Find the area of the base $a=w^2$ Find the area of a side using the height we found above $a=wh$ $a=w\left(\frac{2}{w^{2}}\right)$ $a=\frac{2w}{w^{2}}$ $a=\frac{2}{w}$ Find the total surface area total surface area = base area + $4$(side area) $S=w^2+4\left(\frac{2}{w}\right)$ $S=w^2+\frac{8}{w}$ Re-write as a function $S(w)=w^2+\frac{8}{w}$ Since length can't be negative, and denominators can't be zero, we know $w$ must be greater than zero: $w>0$
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