A farmer wishes to enclose the maximum possible area with 100 meters of fence. The pasture is bordered by a straight cliff, which may be used as part of the fence. What is the maximum area that can be enclosed?

I really don't know the answer to this one, but I assume it has something to do with circles.  My guess would be that the shape would simply be a semicircle, but just in case, I'll widen my optimization to a rectangle connected to the semicircle.

So, picture this:  There is a rectangle bordering the cliff, and the width, the length of the segment of cliff I use, is x meters.  On the opposite side of the cliff, I close it off with a semicircle.

The area as a function of x:
A(x)=pi*x²/8 + (100-pi*x/2)/2 * x
A(x)=pi*x²/8 + 50x - pi*x²/4
A(x)=50x - pi*x²/8
Using what we learned in algebra...
A(x)=-pi/8 * (x² - 400x/pi +40000/pi²) + 5000/pi
A(x)=-pi/8 * (x-200/pi)² + 5000/pi

That means the vertex is at x=200/pi.  That means, that of all the cases I looked over, the semi circle was the best.

It also means that the area at that point is 5000/pi m².  Right now, that's the solution I'm giving.

Posted by Tristan on 2004-06-18 10:07:26