Asked by Gagan

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.
Answered by Reiny
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)
Answered by some one
it 125/ pi and 125 m :D
Answered by Anonymous
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?
Answered by Anonymous
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
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