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)
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?
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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