Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.10 - Modeling with Functions - Exercise Set - Page 293: 32

Answer

The amount of material needed to construct the box in terms of the length of a side of its square base $x$ and a partition down the middle is, $A={{x}^{2}}+\frac{2000}{x}$ square inches

Work Step by Step

The volume of the provided open box $=400\text{ cubic feet}$ Now, use the formula of the volume of the cuboid, $V=l\cdot w\cdot h$ in the above equation $l\cdot w\cdot h=400$ Substitute $x$ for $l$ and $w$ , and $y$ for $h$ in the above equation $\begin{align} & x\cdot x\cdot y=400 \\ & {{x}^{2}}\cdot y=400 \\ \end{align}$ Or $y=\frac{400}{{{x}^{2}}}$ The surface area of given open box $A=lw+2lh+2wh+lh$ Substitute $x$ for $l$ and $w$ , and $y$ for $h$ in the above equation $\begin{align} & A=x\cdot x+2\cdot x\cdot y+2\cdot x\cdot y+x\cdot y \\ & ={{x}^{2}}+5\cdot x\cdot y \end{align}$ Substitute $\frac{400}{{{x}^{2}}}$ for $y$ in the above equation $A={{x}^{2}}+5\cdot x\cdot \frac{400}{{{x}^{2}}}$ Simplify $A={{x}^{2}}+\frac{2000}{x}$ The required amount of material needed to construct the box, $A$ , as a function of the length of a side of its square base $x$ and a partition down the middle is, $A={{x}^{2}}+\frac{2000}{x}$.
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