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Optimization Problem

by pacman99
Tags: optimization
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pacman99
#1
Apr25-04, 04:02 PM
P: 1
Hi,

I just needed help starting off this problem:

"A 1-km racetrack is to be built with two straight sides and semicricles at the ends. Find the dimensions of the track that encloses the maximum area."

There was a similar question which I did before this which involved a Norman window. My idea was, in order to make everything in terms of one variable, let's say the width, I made radius equal to half of the width. The problem is that I DO get an answer but my textbook says that the answer is "a circular track with a radius of 1/2pi km". I find that weird because it says "racetrack is to be built with two straight sides...."

Anyways I just wanted to know, how should I set up the perimeter and area equations? I know the perimeter is 1km but should I solve for y by making r = half of y (width) or should i keep two variables and try to work with them? (width & radius)

Thanks!
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NSX
#2
Apr25-04, 04:51 PM
P: 65
hehe
fun fun

I just did this right now, again; i'm new to the LaTeX, so bare with my text (where there is space with no operators, this means multiply):

I called the radius of the circle r, and the straight part of the track x.

Circumference = C = 2 pi r ; r = radius
Area = A = pi r^2

Total Distance = d = 2 x + 2 pi r = 1 => x = 1/2 - pi r (1)

Area = A = pi r^2 + 2 r x (2)


Sub (1) into (2):

A = pi r^2 + 2 r (1/2 - pi r)

A = r - pi r^2

Take derivative of A w/r.t. r to find when the slope is 0 (since this means max or min):

dA/dr = 1 - 2 pi r

set dA/dt = 0:

1 - 2 pi r = 0 => r = 1 / (2 pi)

Now, you can do the first derivative test, and you see that for r < 1 / (2pi), the slope is positive, and for r > 1 / (2pi), the slope is negative.
Therefore maximum area at a track with a radius of 1 / (2 pi). Furthermore, you can find what the straight portion of the track should be by substituting r into equation (1).

Note that when you substitute r into equation (1), you get:

x = x = 1/2 - pi (1 / (2 pi)

The pis cancel and the 1/2 negate each other, meaning that the x, or straight part of the track should be zero to yield optimal results from the conditions given.

No x, means no straight part, and that means all circle



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