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January 30, 2015

January 30, 2015

Posted by **Mysty** on Thursday, May 19, 2011 at 1:13am.

- Algebra 2 -
**drwls**, Thursday, May 19, 2011 at 6:33amLet L be the length of fence parallel to the river.

A = L*(2400-L)/2 = 1200L - L^2/2

dA/dL = 0 = 1200 - L (using calculus)

L = 1200 m for maximum A

Amax = 1200*600 = 720,000 m^2

You can get the same answer by completing the square.

A = -(1/2)(L^2 -2400L + 1,440,000) + 720,000

= (-1/2)(L-1200)^2 + 720,000

That obviously has a maximum value when L = 1200.

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