Q. 26 - Precalculus Enhanced with Graphing Utilities

Question

An open box with a square base is required to have a volume of 10 cubic feet.

(a) Express the amount A of material used to make such a

box as a function of the length x of a side of the square

base.

(b) How much material is required for a base 1 foot by

1 foot?

(c) How much material is required for a base 2 feet by

2 feet?

(d) Use a graphing utility to graph $A=A\left(x\right)$. For what value

of x is A smallest?

Step-by-Step Solution

Answer

Part (a) Area of material used for constructing open box is $A\left(x\right)=x^2+\frac{40}{x}$.

Part (b) $A\left(1\right)=41i{n}^2$

Part (c) $A\left(2\right)=24i{n}^2$

Part (d) A(x) is smallest when x is $\sqrt[3]{20}$ feet.

Part (a). Step 1. Use the given volume of the open box.

Volume of the open box is

$$\begin{gathered} V=x^{2}h \\ 10=x^{2}h \\ \frac{10}{x^2}=h \end{gathered}$$

Part (a) Step 2. The area of the material required to make the open box is

Area of base $+4$ (area of sides)

$$\begin{gathered} A\left(x\right)=x^2+4xh \\ A\left(x\right)=x^2+4x\left(\frac{10}{x^2}\right) \\ A\left(x\right)=x^2+\frac{40}{x} \end{gathered}$$

Part (b) Step 1. Substitute x = 1 inch in A ( x ) = x 2 + 40 x .

This gives

$$\begin{gathered} A\left(1\right)=1^2+\frac{40}{1} \\ A\left(1\right)=41{\mathrm{in}}^2 \end{gathered}$$

Part (c) Step 1. Substitute x = 2 inches in A ( x ) = x 2 + 40 x .

This gives

$$\begin{gathered} A\left(2\right)=2^2+\frac{40}{2} \\ A\left(2\right)=4+20 \\ A\left(2\right)=24{\mathrm{in}}^2 \end{gathered}$$

Part (d) Step 1. The graph of A ( x ) = x 2 + 40 x is

From the graph,

A(x) is smallest when $x=2.71=\sqrt[3]{20}$ feet.