# Maximum area of rectangle with fixed perimeter.

How can you, with polynomial functions, determine the maximum area of a rectangle with a fixed perimeter.

Here's the exact problem—

You have 28 feet of rabbit-proof fencing to install around your vegetable garden. What are the dimensions of the garden with the largest area?

I've looked around this Stack Exchange and haven't found an answer to this sort of problem (I have, oddly, found a similar one for concave pentagons).

If you can't give me the exact answer, any hints to get the correct answer would be much appreciated.

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The result you need is that for a rectangle with a given perimeter the square has the largest area. So with a perimeter of 28 feet, you can form a square with sides of 7 feet and area of 49 square feet.

This follows since given a positive number $A$ with $xy = A$ the sum $x + y$ is smallest when $x = y = \sqrt{A}$.

You have $2x + 2y = P \implies x + y = P/2$, and you want to find the maximum of the area, $A = xy$.

Since $x + y = P/2 \implies y = P/2 - x$, you substitute to get $A = x(P/2-x) = (P/2)x - x^2$. In your example $P = 28$, so you want to find the maximum of $A = 14x - x^2$.

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So would it be fair to say that the vertex of that function would give the maximum side length and maximum area (x = max side length; y = max area)? –  Ethan Turkeltaub Apr 6 '12 at 22:29
@EthanTurkeltaub Well, no. It gives the maximum area. Maximum side length doesn't really make sense. For a rectangle height and width are not necessarily the same, so there would be two different measurements for side length. Since the perimeter is fixed and we know the perimeter, $28 = 2 * height + 2 * width$ any time you increase the height or the width, you must decrease the other. Also, if you maximize either one, then you would have one of them equal to 14 feet, but that forces the other to be 0 feet (so the total perimeter stays 28ft). –  user23784 Apr 6 '12 at 22:40
@EthanTurkeltaub It would be fair to say that graph shows the area ($y$ on the graph) in terms of either the height or width ($x$ on the graph). As you can see the area is at a maximum when $x = 7$, not when $x$ is at it's maximum possible value of 14. –  user23784 Apr 6 '12 at 22:42
I see what you're saying. That was just a mistake on my part. Thank you very much for your help on this. –  Ethan Turkeltaub Apr 6 '12 at 23:36

Here is a slightly different approach. Let us see what happens if we use a rectangle with base $x$ and height $y$.

Then the perimeter (amount of fencing) used is $2x+2y$. This is $28$, so $2x+2y=28$, or more simply $x+y=14$.

Note that $$4xy=(x+y)^2-(x-y)^2.$$ Since $x+y=14$, it follows that $$4xy=(14)^2-(x-y)^2.$$ To make $4xy$ (and hence $xy$) as large as possible, we must subtract as little as possible from $(14)^2$. So we must make $(x-y)^2$ as small as possible. Since $(x-y)^2$ is a square, it is always $\ge 0$, and it is smallest when $x=y$, that is, when our rectangle is a square.

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+1, very nice way to look at the problem. –  user23784 Apr 6 '12 at 23:03
Ah, that makes a lot of sense. Thank you very much! –  Ethan Turkeltaub Apr 6 '12 at 23:38