# My numbers don't feel right on this maximization question.

If Jack is going to construct a rectangular dog pen and divide it into 3 equal subpens (with two fences inside parallel to one side). What is the maximum area of the overall pen if he only has 240 feet of wire? Remember the wire must be used for the dividers as well as the outsides.

My outcome doesn't seem right at all so I will write each step of my thought process out.

So I know that the area will be:

$A = xy$

My equation should look like:

$C = 4x + 2y$

So then I want to single out a variable:

$240 = 4x + 2y$

Divide by 2:

$120 = 2x + y$

This will give us the y-value:

$y = \frac{120}{2x}$

Lets plug that into the equation above:

$C = 4x + 2(\frac{120}{2x})$

Which will leave us with:

$C = 4x + \frac{120}{x}$

Derivative of that is:

$C' = -\frac{120}{x^2} + 4$

Set that equal to 0:

$-4 = -\frac{120}{x^2}$

Divide by -1 and multiply by $x^2$:

$4x^2 = 120$

Simplify:

$x = \sqrt{30}$

Plug that back in to my original equation:

$240 = 4(\sqrt{30}) + 2y$

Which leaves my y-value at:

$\frac{240}{2*4*\sqrt{30}}$

If I plug my anwers back in they don't add up. What did I do wrong?

EDIT

Reworked with the mistake fixed:

$y = 120 - 2x$

Plug into equation:

$A = x * (120 - 2x)$

Simplified:

$A = 120x - 2x^2$

Take derivative:

$A' = 120 - 4x$

X value:

$x = 30$

Find y-value:

$120 - 2(30) = 60 Get Area:$ 30 * 60 = 1800

That seems much better!

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You used $y = \frac{120}{2x}$, but really $y = 120 - 2x$. – hardmath Dec 12 '12 at 18:34
@hardmath Thank you for pointing that out! Can't believe I made such a stupid mistake. – StrugglingWithMath Dec 12 '12 at 18:51

When you plug $y=\frac {120}x$ it should be $120-2x$ and you should plug it into the equation for $A$, as that is what you want to maximize, then take $\frac {dA}{dx}$. A resonableness check is that you have six sides, so they should be something like $\frac {240}6=40$ feet long.