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.
$\endgroup$4 Answers
$\begingroup$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 $A = 14x - x^2$.
$\endgroup$ 4 $\begingroup$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.
$\endgroup$ 5 $\begingroup$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.
$\endgroup$ 5 $\begingroup$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.
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