# Thread: Maximum Area of a Rectangle

1. ## Maximum Area of a Rectangle

Question: One-hundred (100) feet of fence is used to fence in a rectangular yard. Find the length and width that give maximum area.

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$\displaystyle 2x + 2y = 100$
$\displaystyle x + y = 50$

Formula for the area is, $\displaystyle A = xy$.
As $\displaystyle x +y = 50$, $\displaystyle y = 50 -x$
So our formula becomes, $\displaystyle A = x (50-x) = 50x - x^2$.

To find the extremum point, we will differentiate it and $\displaystyle A' = 0$ will give us the extremum point.

$\displaystyle A = 50x - x^2$
$\displaystyle A' = 50 - 2x$
$\displaystyle A' = 50 - 2x = 0$
$\displaystyle x = 25$

There's an extremum point at $\displaystyle x = 25$, but we don't know whether it's a minima or a maxima. The second derivative test will help us to determine what it is

If $\displaystyle A''(25) > 0$, then it's a minima. If $\displaystyle A''(25) < 0$, it's a maxima.

$\displaystyle A = 50 x - x^2$
$\displaystyle A' = 50 - 2x$
$\displaystyle A'' = -2$
$\displaystyle A''(25) = -2$, which is less than 0 and which makes our point a maxima

So $\displaystyle x = 25$ will give us the maximum area.

3. x+y=50
so
50=x+y>=2*sqrt(x*y) (when x=y then x+y=sqrt(x*y))
so we have x*y<=25^2=625 (when x=y=25)

Calculus is unnecessary

4. Originally Posted by invite_moon
x+y=50
so
50=x+y>=2*sqrt(x*y) (when x=y then x+y=sqrt(x*y))
so we have x*y<=25^2=625 (when x=y=25)
Your solution is OK, but can you apply it to more complicated functions?

Originally Posted by invite_moon
Calculus is unnecessary
This question is posted under the Calculus forum, so I treated it like any optimization question.