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November 23, 2014

November 23, 2014

Posted by **Gagan** on Tuesday, March 24, 2009 at 6:52pm.

can somebody plz help me this is really important and any help will be greatly appricieted.

thnx a lot.

- math -
**Reiny**, Tuesday, March 24, 2009 at 7:53pmlet 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)

- math -
**some one**, Monday, January 17, 2011 at 8:18pmit 125/ pi and 125 m :D

- math -
**Anonymous**, Thursday, January 27, 2011 at 2:16pmthe 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?

- math -
**Anonymous**, Monday, October 17, 2011 at 8:46pmgrade 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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