Min/Max Calculus Problems

**aussiekid90** · January 9th 2007, 03:44 PM

Hi, i have two problems of bounded areas which i just don't even know how to start. If you could give me some pointers on how to do bounded area problems in general as well, that would be awesome.

A rectangle is bounded by the x-axis and the semicircle
y= squareroot(25-x^2) what length and wideth should the rectange have so that its are is maximum?

A rectangular package to be sent by a postal service can have a amazimum combined length and girth of 108 inches. Find the dimensions of the package of maximum volume that can be sent.

Thanks a lot.

Aussiekid90

**ThePerfectHacker** · January 9th 2007, 04:34 PM

Originally Posted by aussiekid90

Hi, i have two problems of bounded areas which i just don't even know how to start. If you could give me some pointers on how to do bounded area problems in general as well, that would be awesome.

A rectangle is bounded by the x-axis and the semicircle
y= squareroot(25-x^2) what length and wideth should the rectange have so that its are is maximum?

Maximum in area?
(The problem probably assumes symettric in the Region).

Let the rectangle be moved over $x$ units the origin. That means the length is $2x$ and the height until the semicircle is $f(x)=\sqrt{25-x^2}$ .
Thus, the area it has is,
$A(x)=2x\sqrt{25-x^2}$
$A'(x)=(2x)'\sqrt{25-x^2}+(2x)(\sqrt{25-x^2})'$
$A'(x)=2\sqrt{25-x^2}+(2x)\left( -\frac{x}{\sqrt{25-x^2}} \right)=0$
Multiply through by denominator,
$2(25-x^2)+2x(-x)=0$
$50-2x^2-2x^2=0$
$4x^2=50$
$x=\pm \frac{\sqrt{50}}{2}$
Note the plus-mius does not matter all it says we can start drawing the rectangle from either side.

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