Calculus question

Hey guys, sorry to hit you with a homework thread, but I'm trying to help out my sister in college.  She's asking for help with this question and I've never taken a calculus class so I'm not totally clear on how to get the answer:

A sports field is to have the shape of a rectangle with semi-circles put on the two ends. It must have a perimeter of 1000 meters. What is the maximum area possible for the rectangular part.

I pretty much found the answer here http://www.ltcconline.net/greenl/courses/103a/keys/PracticeFinal/practiceFinaKeyl.htm on problem 11, but I don't know what "this can also be done by hand using the vertex of a parabola formula" means.  The rest of it makes sense to me.

Thanks for any help.

OK, the given information is that rectangle area A = x2r and x = 600 - pr where p is pi.  Rearrange the second equation to get r = (600-x)/p, then subsitute (600-x)/p for r in the area equation, to get A = 2x(600-x)/p = (1200x - x²)/p.  So Ap = -x[sup]2 + 1200x.  You should recognize that you've got an equation describing a parabola oriented so that its vertex is at the maximum.  You might have a formula available that finds the vertex based on A, B, and C values for an equation of the form Ax² + Bx + C = 0.

And for y = -x² + 1200x - Ap = y (I should have put  "y =" in post >>2 and not "= 0", unless you wanted to solve for x-intercepts), A = -1, B = 1200, and C = -Ap.

Suppose the rectangle is w-by-h and the semi-circles are connected to the h edges. We want to maximize the rectangle area, wh, given that the perimeter of the whole field is exactly 1000, i.e. 2w + h\pi = 1000. Solving this equation for h gives us h in terms of w: h = \frac{1000-2w}{\pi}. We can then write the rectangle area in terms of only w: wh = \frac{w(1000-2w)}{\pi}. Plugging graph w(1000-2w)/pi into www.wolframalpha.com, we can see the graph of the rectangle area for various values of w. Notice that it is a parabola with its peak somewhere between w = 200 and w = 300.

In high school, we learn how to find the exact w value of the peak by rewriting \frac{w(1000-2w)}{\pi} in a specific form. In her Calculus class, your sister learns to find it by finding where the derivative is zero. She must find the derivative of \frac{w(1000-2w)}{\pi} by hand, then figure out what value of w makes it zero.

Being lazy, we can get www.wolframalpha.com to do it for us by inputting differentiate w(1000-2w)/pi. It tells us the derivative and that it is zero at w = 250. This is exactly where the peak of the parabola is. So w = 250 maximizes the rectangle area, and we can get our final answer: the maximum rectangle area is \frac{250(1000-2(250))}{\pi}, which is about 39789.

>>1
Tell her to read the fucking book.

http://www.anontalk.se/