math
posted by Gagan on .
A highschool is planning to build a new playing field surrounded by a running track. The track coach wants two laps around the track to be 1000m. The football coach wants the rectangular infield area to be as large as possible. Can both coaches be satisfied?
can somebody plz help me this is really important and any help will be greatly appricieted.
thnx a lot.

let the length of the rectangular infield be x m
let the width of the rectangular infield be 2r m ,(making the radius of the semicircular ends r m)
once around is 500 m
2x + 2pir = 500
x + pir = 250 or x = 250  pir
Area of infield = x(2r)
= 2r(250  pir)
= 500r  2pir^2
dA/dr = 500  4pir = 0 for a max of A
r = 39.788
x = 250  pi(39.788)
= 125
so the length of the infield should be 125 m
and its width should be 79.58 m
(Wow, what a field) 
it 125/ pi and 125 m :D

the answer to this question can not use derivatives (or shouldn't) because it is in a grade 11 math textbook within a quadratics unit. Anyone else want to give it a go?

grade 11:
2pir + 2w = 500m
w = 250  pir
Area of in = x(2r)
= (250  pir)2r
it's a "partially factored" quadratic...
0 = (250  pir)2r
r = 0, 250/pi
(0 + (250/pi))/2 = 125/pi
r = 125/pi
...
w = 125
250/pi m by 125 m