## Precalculus (6th Edition) Blitzer

The amount of material needed to construct the box in terms of the length of a side of its square base $x$ is, $A={{x}^{2}}+\frac{40}{x}$ square feet.
Volume of the provided open box $=10\text{ cubic feet}$ Now, use the formula of the volume of cuboid, $V=l\cdot w\cdot h$ in the above equation $l\cdot w\cdot h=10$ Substitute $x$ for $l$ and $w$ , and $y$ for $h$ in the above equation \begin{align} & x\cdot x\cdot y=10 \\ & {{x}^{2}}\cdot y=10 \end{align} Or $y=\frac{10}{{{x}^{2}}}$ The surface area of given open box $A=lw+2lh+2wh$ 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}^{2}}+4\cdot x\cdot y \end{align} Substitute $\frac{10}{{{x}^{2}}}$ for $y$ in the above equation $A={{x}^{2}}+4\cdot x\cdot \frac{10}{{{x}^{2}}}$ Simplify $A={{x}^{2}}+\frac{40}{x}$ The required amount of material needed to construct the box in terms of the length of a side of its square base $x$ is, $A={{x}^{2}}+\frac{40}{x}$ square feet.