Question

$24 is available to make a rectangle storage container with a square base and an open top. If the material for the base cost $8 per square meter and the material for the side costs $2 per square meter, find the dimensions of the container with the largest possible volume

Answer 1

First, draw a diagram of the box identifying two relevant variables: the side length of the square base and the height of the box:

Find an expression of the cost of the container in terms of L and h. To do so, find first the lateral surface area and the area of the base, multiply each area times its cost per square meter and add them.

The area of the base is given by the expression:

`L^2`

The lateral surface area is four times the area of one rectangular side. The area of a single rectangular side is given by the expression:

`hL`

Then, the lateral surface area is:

`4hL`

Since the cost of the material is $2 per square meter on the sides and $8 per square meter on the base, then, the total cost of the rectangular box, is:

`\begin{gathered} C=8* L^2+2*4hL \\ =8L^2+8hL \end{gathered}`

Since $24 is available to make the container, substitute C=24:

`\begin{gathered} \Rightarrow24=8L^2+8hL \\ \Rightarrow(24)/(8)=(8L^2+8hL)/(8) \\ \Rightarrow3=L^2+hL \end{gathered}`

Isolate h from the equation:

`\Rightarrow h=(3-L^2)/(L)`

The volume of the container is equal to the area of the base times its height:

`V=L^2h`

Substitute the expression for h to find the volume only in terms of L:

`\begin{gathered} V=L^2*(3-L^2)/(L) \\ =L(3-L^2) \\ =3L-L^3 \end{gathered}`

For the volume to be the maximum possible, then the value of L must be such that the expression 3L-L^3 has the maximum value (with the consideration that both volume and length must be positive).

Graphically, we can see that this happens when L=1. The graph of the expression 3L-L^3 is shown:

Substitute L=1 in the equation for h to find the value of h for the maximum volume configuration:

`\begin{gathered} h=(3-L)/(L^2) \\ =(3-1)/(1^2) \\ =(2)/(1) \\ =1 \end{gathered}`

Therefore, the dimensions of the container with the maximum possible volume such that the cost is equal to $24, are:

`\begin{gathered} \text{ Height: 2 meters} \\ \text{Side length of the base: 1 meter} \\ \text{Volume: 2 cubic meters} \end{gathered}`