Question

A rectangular storage container with an open top is to have a volume of $10{\mathrm{m}}^3$.

The length of this base is twice the width.

Material for the base costs $\$5\mathrm{per}\mathrm{square}\mathrm{meter}$ .

Material for the sides costs $\$3\mathrm{per}\mathrm{square}\mathrm{meter}$ .

Find the cost of materials for the cheapest such container. (Round your answer to the nearest cent.)

Answer 1

Find the cost of material to make cheapest container:

It is given that the length of the base of container is twice of width.

Let the width of the base of container is $x\mathrm{meter}$.

Then the length of the base of container is $2x\mathrm{meter}$.

It is given that the volume of container is $10{\mathrm{m}}^3$.

So that the height of the container is given by :

$\begin{array}{rcl}h& =& \frac{10}{2x\times x}\\ & \Rightarrow & h=\frac{5}{x^2}\end{array}$

The cost of base of container is $\$5{\mathrm{m}}^2$.

So that the total cost of base is : $2x\times x\times 5=\$10x^2$.

Also the cost of sides is $\$3{\mathrm{m}}^2$.

So that the total cost of sides of the container is $\begin{array}{rcl}& =& \left[2\left(2x\times \frac{5}{x^2}\right)+2\left(x\times \frac{5}{x^2}\right)\right]\times 3\end{array}$

$\begin{array}{rcl}& =& \left[\frac{20}{x}+\frac{10}{x}\right]\times 3\\ & =& \left[\frac{30}{x}\right]\times 3\\ & =& \$\frac{90}{x}\end{array}$

So that the function for total cost of container is :-

$f\left(x\right)=10x^2+\frac{90}{x}$

To minimize the function differentiate it and take differential equals to zero. Then :

$$\begin{gathered} f\text{'}\left(x\right)=20x-\frac{90}{x^2}=0 \\ \Rightarrow 20x^3-90=0 \\ \Rightarrow 20x^3=90 \\ \Rightarrow x=1.65 \end{gathered}$$

So that minimum value of the function is :-

$\begin{array}{rcl}f\left(1.65\right)& =& 10{\left(1.65\right)}^2+\frac{150}{1.65}\\ & \Rightarrow & f\left(1.65\right)=81.77\end{array}$

Hence, so that the cost of cheapest rectangular container is $\mathbf{\$}\mathbf{81}\mathbf{.}\mathbf{77}$.