Christopher F. answered • 04/21/20

M.S. in Mathematics with 500+ hours in Calculus tutoring

The volume of the box with square base can be written as V = s^{2 }h, which is 1053 ft^{3}.

So we have 1053 = s^{2 }h, representing our constraint equation.

Let's set up the total cost function:

The base of the box is s^{2} ft^{2 }which costs $8 per ft^{2}

The roof of the box is also s^{2} ft^{2 }but costs $5 per ft^{2}

Lastly, the sides of the box total 4(s*h) ft^{2} and cost $4.50 per ft^{2}

Multiplying and summing, our total cost function looks something like this: C = 8s^{2 }+ 5s^{2} + 4.50*4(s*h)

C = 13s^{2 }+ 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/s^{2} which we'll plug into C:

C = 13s^{2 }+ 18s*h = 13s^{2 }+ 18s*(1053/s^{2}) = 13s^{2 }+ 18(1053)s^{-1 }= 13s^{2} + 18954s^{-1}

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

Set C' equal to 0: 26s - 18954/s^{2} = 0 ----> 26s=18954/s^{2} ----> (multiply both sides by s^{2}) 26s^{3} = 18954

----> s^{3} = 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. :)