Q. 43 - Calculus

Question

For Jen, Alina wants to make a box with a square base whose sides and base are made of wood and whose top is made of metal. The wood she wants to use costs $5$ cents per square inch, while the material for the metal top costs $12$ cents per square inch. What is the largest possible box that Alina can make for Jen if she only has $\$20.00$ to spend on materials?

Step-by-Step Solution

Answer

Ans:  The dimensions of the largest possible box are $6.26\times 6.26\times 10.65$   

         The largest possible box is $\text{ 417.48 cubic inches }$

Step 1. Given information.

given,

  The figure above shows a cube of side $l$ inches.

  The cost of the metal top is $\$0.12$ per square inch and the cost of the remaining faces is $\$0.05$ per square inch.

Step 2. The objective is to find the largest dimensions of the box such that the cost of the box is within $ 20 .

The area of each of the four sides face is,

       $l h$

Therefore, the cost of making a side face is, 

      $0.05 lh$

The cost of making $4$ side faces is,

    $40.05=0.2\mathrm{lh}$

The cost of making the top is,

        $0.12l^2$

Cost of making bottom is,

        $0.05l^2$

Therefore, total cost of making the box is,

       $0.05l^2+0.12l^2+0.2lh$

Consider the cost should is $\$20$,
 Therefore,

       $\begin{array}{r}0.05l^2+0.12l^2+0.2lh=20\\ 0.17l^2+0.2lh=20\end{array}$

Step 3. The volume of the box is given by,

     $V=l^{2}h$

Substitute $l h$ from the cost relation,

     $\begin{array}{r}V\left(l\right)=l\left(20-0.17l^2\right)\\ =20l-0.17l^3\end{array}$

differentiate the volume with respect to $l$,

     $V^{\mathrm{\prime }}\left(l\right)=20-0.51l^2$

Equate the derivative to zero and solve for $l$,

     $\begin{array}{r}20-0.51l^2=0\\ l^2=\frac{20}{0.51}\\ l\approx 6.26\end{array}$

Therefore, $h$ is,

      $\begin{array}{r}h=\frac{20-0.17(6.26{)}^2}{0.2\left(6.26\right)}\\ \approx 10.65\end{array}$

Therefore, the dimensions of the largest possible box are $6.26\times 6.26\times 10.65$.

The maximum volume is given by,

    $\begin{array}{r}V=\left(6.26{)}^2\right(10.65)\\ \approx 417.48\end{array}$

The maximum volume is $417.48\text{ cubic inches }$ cubic inches.