Maximum Area of a Rectangle

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February 4th 2008, 01:37 PM
currypuff

Maximum 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.
February 4th 2008, 03:23 PM
wingless

Code:

_______x_______ | | | | y| |y | | |_______________| x

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

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

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

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

There's an extremum point at $x = 25$ , 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 $A''(25) > 0$ , then it's a minima. If $A''(25) < 0$ , it's a maxima.

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

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