A shelter at a bus stop is ti be made with three Plexiglas sides and a Plexiglas top.

Question

I don't have the shape or drawing unfortunately. 

If the volume of the shelter is 486 cubic ft, find the dimensions that require the least amount of Plexiglas.

Answer 1

I would start by saying that the shelter's volume and surface area could be modeled by equations like

 

Volume = 486 = LWH

 

and 

 

Surface area = 2WH + LW + LH

 

Is there anymore information given about how two of these dimensions are related? Does it say that two of the sides are squares or that the length is twice the width or something like that? Because the problem is that we have three unknowns and only two equations. 

 

Maybe this is possible if you're in Calc 3, hang on...

 

If I call surface area S(LWH) = 2WH + LW + LH I can finagle this to have two variables with appropriate substitutions.

 

Since I have  486 = LWH I can say that LW is 486/H and LH is 486/W. I can then substitute two terms in S to get

 

S(L, W, H) = 2WH + 486/H + 486/W = S(W, H)

 

Then if I take the partial derivatives of S with respect to W and H, I can get that

 

S_W = 2H - 486/W²           and

 

S_H = 2W - 486/H²

 

Any H and W that will be a minimum will be found at critical points of S, so we consider the following system of equations: 

 

2H - 486/W² = 0

 

2W - 486/H² = 0

 

Then we get that 2W = 486/H² and that WH² = 243

 

Considering the first equation in the system, I can rewrite it as 2H = 486/W² or equivalently, W²H = 243.

 

Considering both  WH² = 243 and W²H = 243, we have that WH² = W²H and can rewrite this and factor it as we do below:

 

 

 

WH² - W²H = WH(H - W) = 0. 

 

Now, all along, we've been assuming that none of the dimensions are zero. Thus, H = W.

 

Knowing that two of our dimensions must be equal, we can return to the original set of equations for the volume and surface area and rewrite them in a more solvable form:

 

 

Volume = 486 = LWH = LW²

 

 

Surface area = 2WH + LW + LH = 2W² + 2LW

 

 

I'm going to use the given volume to substitute for L: L = 486/W²

 

Now I can get the surface area in terms of just W: S = 2W² + 972/W

 

To find W that will make this minimal (hopefully minimal) we take the derivative of S with respect to W to get

 

dS/dW = 4W - 972/W² = 0             (and set it equal to zero to find critical points)

 

Then W³ - 243 = 0

 

Which makes W about 6.24

 

Now that we know W, we know H is the same thing and we can use L = 486/W² to find L.

 

Then L is about 12.48.

 

So the dimensions should be 6.24, 6.24, and 12.48 thereabouts. 

 

Unless I effed something up, which is totally possible.