Derivitive Application - Playing field perimeter/area

[ivanwind15]

Alright, here is the problem:

A playing field is to be built in the shape of a rectangle plus a semicircular region at each end. A 400 meter racetrack is to form the perimeter of the field. What dimensions will give the rectangular part the largest area?

So I drew a rectangle with two semicircles at each end, and came up with the equation:

400 = 2?r + 2y

Tried solving for y, and got:

(400 - 2?r) / 2 = y Which equals 200 - ?r

Put it into the area equation and got:

A = ?r^2 + (200 - ?r) * 2r which simplifies to 400r - ?r^2

Took the derivative so...

0 = 400*dr/dt - 2?r*dr/dt

But, now I am stuck here. I don't know how to find r or dr/dt.

Thank you for the help.

 

[skeeter]

you want to maximize the rectangular part of the field ...

let A = the area of the rectangular part of the field.
y = length of the rectangular part of the field
2r = width of the rectangular part of the field

\(\displaystyle A = 2ry\) where \(\displaystyle y = 200 - \pi r\)

also, why are you taking the derivative w/r to time, t ???

now find \(\displaystyle \frac{dA}{dr}\) and maximize.