# Maximum Area of a Rectangle

• Feb 4th 2008, 01:37 PM
currypuff
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.

• Feb 4th 2008, 03:23 PM
wingless
Code:

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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.
• Feb 4th 2008, 06:00 PM
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)

Calculus is unnecessary
• Feb 4th 2008, 11:06 PM
wingless
Quote:

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?

Quote:

Originally Posted by invite_moon
Calculus is unnecessary

This question is posted under the Calculus forum, so I treated it like any optimization question.