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I am stuck on another optimization problem and I can't get my answer to match the author's. I am assuming the author is correct, but there is no justification for their answer.

A farmer with 750 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens?

I know that the two formulas I need are

$2w + 5l = 750$ w for width and l for length to find the perimeter. I know it is this because there are 3 fences laid down inside a rectangle, so that gives me 5 lines of fence in one way and then 2 in the other direction.

$4lw = a$ for this I know that I have 4 rectangles so it is 4 lw.

Subsituting in w in terms of l

$2l(750-5l) = a$

$1500l-10l^2=a$

then I take the derivative

$1500 - 20l = a$

$l=75$ which gives me the wrong answer.

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    @Austin I don't understand, I set up the problem to be 5w and 2l so the cooresponding area would use the same shapes. I suppose it would be 1/4w and 4l though.2012-04-04

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Your mistake seems to be that you're confusing the width of each pen with the width of the whole area. The formula $2w + 5l = 750$ assumes that $w$ is the width of all four pens added together, yet $4lw = a$ treats $w$ as the width of only one pen. Once you make the definition of $w$ consistent, you should get the right answer.

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Always draw a picture. The pens should look something like this: enter image description here

Total length of fencing: $5l+2w=750 \mbox{ ft}$

$Area=l\times w =\frac{750}{2}l-\frac{5}{2}l^2\mbox{ ft}^2$

To find the value of $l$ that maximises the area we differentiate the area formula above with respect to $l$ and set it to zero, then solve for $l$

$\frac{dArea}{dl}=\frac{750}{2}-5l=0$

which has solution $l=75 \mbox{ft}$ and so the required maximum area is $ Area_{max}=\frac{750}{2}75-\frac{5}{2}75^2=14062.5\mbox{ ft}^2 $ (that this is a maximum you can confirm by taking the second derivative of the area with respect to $l$ and examining its sign)