# Hi I need help with this optimization problem for calculus. In the wake of a natural disaster, a civil engineer is tasked with building a one room structure, in the form of a rectangular prism....

Hi I need help with this optimization problem for calculus.

In the wake of a natural disaster, a civil engineer is tasked with building a one room structure, in the form of a rectangular prism. The structure must have a height of 10 feet in order to fit the required equipment to be used in the room. Of course, there are limited resources. In particular, the civil engineer has only enough drywall to cover walls with a total surface area of 2000 square feet. Determine the dimensions of the room that give a maximum volume.

I need this question solved ASAP.

*print*Print*list*Cite

Hello!

Let's set up the problem with the variables. Besides the given height `H = 10 ft,` a rectangular prism also has a length `L` and a width `W` (both in feet). Of course we know that the volume of the rectangular prism is `V = H*L*W ft^3 = 10L*W ft^3,` and we need to maximize it.

Next, the constraint relating to the walls surface. It is not clear whether "walls" means four side walls excluding both the ceiling and the floor. I suppose walls are four vertical walls only. There are two walls of the dimensions `L xx H` and two walls `W xx H,` so their total area is `2*L*H + 2*W*H,` and this is equal to `S = 2000 ft^2.`

This equation, `2*L*H + 2*W*H = 2*L*10 + 2*W*10 = 2000,` gives us the simple constraint `L + W = 2000/20 = 100,` so `L = 100 - W.` The function we need to maximize becomes `V = V(W) = 10(100-W)W = -10W^2 + 1000W.`

It is the quadratic function of `W` with the negative factor `a = -10` at `W^2` and the factor `b=1000` at `W^1.` Its graph is a parabola branches down and it has the only maximum. The point where it is reached is `W_0 = -b/(2a) = -1000/(-20) = 50 (ft).`

Thus `L_0 = 100 - W_0 = 50 (ft),` and the volume they give is `V = 10*50*50 = 25000 (ft^3).`

The answer: the (horizontal) dimensions of the room that give a maximum volume are **50 ft** and **50 ft**. The maximum volume is `25000 ft^3.`

(if we have to take the ceiling and/or the floor with the area `L*W` each into account, the function remains quadratic and the solution remains similar)