Friday, May 2, 2008

Optimization problems

Today, we were doing optimization problems.
In order to describe how we are solving these problems, I will be using an example from the slide 5.
It is the open box question where you have to cut squares from the corners to maximize the volume of the box.

First step is to draw a picture of what you want to do. We are cutting squares from the corners and we label the square's side with x. We know that the side of the entire piece is 24 inches per side. Therefore the equation is (24-2x) for the space between the two corner squares you have to cut. Everything is a square so all sides are the same.

Next we know that the volume of the box is going to be V=L*W*H.
We know that the height(H) is x and each side (L & W) is the same and equal to (24-2x), so the equation is going to be
V(x)= (24-2x)*(24-2x)*x
which is
V(x)=((24-2x)^2) * x

Third step is to simplify the equation and you will get
V(x)=(4x^3) - (96x^2) + 576x.

We take the derivative of that to find the max.
V'(x) =(12x^2) - 192x + 576
V'(x) =12((x^2) - 16x + 48)

The derivative at the max is equal to zero.
So we let V'(x) = 0
0 = 12((x^2) - 16x + 48)
0 = ((x^2) - 16x + 48)
0 = (x - 4) (x - 12)
x = 4, 12
x can't equal to 12 because that would make the side of the box equal to zero.
Therefore x = 4.

The last step is to test it.

4 by 4 squares should be cut out to maximize the volume of the box.

The next scribe is Dino.

No comments:

Post a Comment