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I was looking through some math problems I came across one where it asks you to prove that to maximize the area of a $3$-sided rectangle the length of the fence(parallel to the river) must be twice the width of the fence(perpendicular to the fence). I believe you are supposed to use quadratics in some form to solve the question. This is where I reached (but I think it is a dead end)...

Let $L$ represent the length of the fence in m. Let $W$ represent the width of the fence in m.Then, $P= L + 2W $ and so $L = P-2W$. Thus, $$A = LW=(P-2W)W=-2W^2 + PW.$$ I do not know where to go from here! Any help would be appreciated! Thanks!

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You need to pick $W$ to maximize $$ A = PW-2W^2.$$

Use the completing the square trick to write: $$ A = -2(W-P/4)^2 +\frac{P^2}{8} $$

Since $P$ is fixed, the only term you have control over is the first term, which is always negative cause of the square. So to make the expression maximum, you want that negative term to be as small in absolute value as possible. The best you can do is set $W=P/4$ so that it's zero.

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    Thank you so much @spaceisdarkgreen. – 2017-01-11
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Another way to find the maximum is use AM-GM:

$$\frac{P}{2}=\frac{L+2W}{2}\ge \sqrt{2LW}=\sqrt{2A} \rightarrow A \le \frac{P^2}{8}$$

So, the maximum is

$$A_{max}=\frac{P^2}{8}$$

and it happens when $L=2W$.