Optimize Real World Solutions: Storage Shed

Question

A storage shed is to be built in the shape of a box with a square base. It is to have a volume of 1053 cubic feet. The concrete for the base costs ​$8 per square​ foot, the material for the roof costs ​$5 per square​ foot, and the material for the sides costs ​$4.50 per square foot. Find the dimensions of the most economical shed.

Answer 1

The volume of the box with square base can be written as V = s² h, which is 1053 ft³.

So we have 1053 = s² h, representing our constraint equation.

Let's set up the total cost function:

The base of the box is s² ft² which costs $8 per ft²

The roof of the box is also s² ft² but costs $5 per ft²

Lastly, the sides of the box total 4(s*h) ft² and cost $4.50 per ft²

Multiplying and summing, our total cost function looks something like this: C = 8s² + 5s² + 4.50*4(s*h)

C = 13s² + 18(s*h)

Now, we want to minimize the total cost, so we'll need to take the derivative of our total cost function. We cannot do that yet because there are two variables in the total cost function. If we solve for one of the variables in our constraint equation and substitute that variable into our total cost function, we'll have a function of one variable, allowing us to the derivative, set equal to zero, and solve.

From the constraint equation, h = 1053/s² which we'll plug into C:

C = 13s² + 18s*h = 13s² + 18s*(1053/s²) = 13s² + 18(1053)s^-1 = 13s² + 18954s^-1

So then, C' = 26s - 18954s^-2

Set C' equal to 0: 26s - 18954/s² = 0 ----> 26s=18954/s² ----> (multiply both sides by s²) 26s³ = 18954

----> s³ = 729 ----> s = 9

Plug s = 9 back into the constraint equation to solve for h: 1053 = 81h ----> h=13

Therefore, s = 9ft and h = 13ft.

Hope this helps! Please comment below if you have any questions about the steps. :)