Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 6 down vote accepted

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 $A = 14x - x^2$.

share|cite|improve this answer
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

Does this help?

Edit: Also, is it elsewhere in the problem that it has to be a rectangle? Because, otherwise a rectangle would not be the best choice.

share|cite|improve this answer
Yes, it must be a rectangle. – Ethan Turkeltaub Apr 6 '12 at 21:48
I'm confused as to why you used 14 instead of 28. – Ethan Turkeltaub Apr 6 '12 at 21:49
because to get the area, you multiply length times width. If you only have 28 feet longest a side could be would be 14 – Kevin Apr 6 '12 at 21:51
So, in this instance, the answer would be the vertex of the parabola (i.e., x = max side length; y = max area)? – Ethan Turkeltaub Apr 6 '12 at 21:53
Yep. Although I must say that rar's is the better answer in that it gives much better detail. – Kevin Apr 6 '12 at 22:00

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.

share|cite|improve this answer
+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
Hello, could you expand on how you got to the 4xy= (x+y)2-(x-y)2 – captainjamie Mar 18 '15 at 9:47
@captainjamie: Recall that $(x+y)^2=x^2+2xy+y^2$ and $(x-y)^2=x^2-2xy+y^2$. Subtract. – André Nicolas Mar 18 '15 at 15:04
@André ok so you're not coming from some fundamental definition of areas and perimeters or anything, just stating a relationship between x and y. I just wondered if you'd got it from differentiating or something. Thanks. – captainjamie Mar 18 '15 at 15:48

Put the perimeter into the vertex formula.

(Find the P=a+b equation and put it into the A=a*b equation.)

The answer is the maximum point.

share|cite|improve this answer

protected by Zev Chonoles Feb 12 at 9:43

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.