Please help me solve this problem:

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

Thanks in advance for your assistance.

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- Feb 4th 2008, 01:37 PMcurrypuffMaximum Area of a Rectangle
Please help me solve this problem:

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

Thanks in advance for your assistance. - Feb 4th 2008, 03:23 PMwinglessCode:
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$\displaystyle x + y = 50$

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. - Feb 4th 2008, 06:00 PMinvite_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)

Calculus is unnecessary - Feb 4th 2008, 11:06 PMwingless