Wednesday

April 16, 2014

April 16, 2014

Posted by **jeremy** on Thursday, March 14, 2013 at 7:24am.

- precalculus -
**Reiny**, Thursday, March 14, 2013 at 8:14amlet the width of the whole rectangle be x m (there will be 4 of these)

let the length be y m

then 4x + 2y = 1200

2x + y = 600

y = 600 - 2x

Area = xy

= x(600-2x)

= -2x^2 + 600x

Now, I don't know if you are studying Calculus.

If you do, then

d(Area)/dx = -4x + 600

= 0 for a max area

x = 150

then y = 600 - 2(150) = 300

and the max area is (150)(300) = 45000

If you don't know Calculus, complete the square on the above quadratic

you should end up with

Area = -2(x-150)^2 + 45000

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