Originally Posted by

**beanus** I think I have this one right but I'm not sure. Here's what I have

The problem states: A farmer plans to fence in a rectangular pasture containing three adjacent pens of equal size, lined up end-to-end. The total length of fencing needed to enclose the three pens is 1200 yards. Find the maximum total area that all three pens can enclose.

So I set up the equation as follows

$\displaystyle A=3xy$

$\displaystyle 6x+4y=1200$

$\displaystyle y=\frac{-6x+1200}{4}$

$\displaystyle A=3x(\frac{-6x+1200}{4})=\frac{18x^2+3600x}{4}$

Apply Quotient Rule

$\displaystyle A'=\frac{-144x+14400}{16}=-9x+90$