# Find Dimensions of Rectangle

• Nov 8th 2008, 08:49 AM
magentarita
Find Dimensions of Rectangle
A special window in the shape of a rectangle with semicircles at each end is to be constructed so that the outside dimensions are 100 feet in length. Find the dimensions of the rectangle that maximizes its area.

• Nov 8th 2008, 09:04 AM
galactus
You need to know the formulas for the circumference and area of a circle.

The circumference of a semicircle is $C={\pi}r$

But, since there are two semicircles that make one whole circle, we have:

$C=2{\pi}r$

But, the radius of the circle is w/2.

$C=2{\pi}(\frac{W}{2})$

The perimeter of the rectangular portion is easy enough with just 2L.

So, the total circumference must be 100 and is:

$2{\pi}(\frac{W}{2})+2L=100$......[1]

Now, for the area:

The area of the entire region is $A={\pi}(\frac{W}{2})^{2}+LW$

Since we must maximize area, solve [1] for L or W (whatever you want or whatever is easiest) and sub into the area equation. It will then be entirely in terms of one variable which you can differentiate and find the solution to the problem.
• Nov 8th 2008, 10:13 PM
magentarita
Wonderfully done!
Quote:

Originally Posted by galactus
You need to know the formulas for the circumference and area of a circle.

The circumference of a semicircle is $C={\pi}r$

But, since there are two semicircles that make one whole circle, we have:

$C=2{\pi}r$

But, the radius of the circle is w/2.

$C=2{\pi}(\frac{W}{2})$

The perimeter of the rectangular portion is easy enough with just 2L.

So, the total circumference must be 100 and is:

$2{\pi}(\frac{W}{2})+2L=100$......[1]

Now, for the area:

The area of the entire region is $A={\pi}(\frac{W}{2})^{2}+LW$

Since we must maximize area, solve [1] for L or W (whatever you want or whatever is easiest) and sub into the area equation. It will then be entirely in terms of one variable which you can differentiate and find the solution to the problem.

Wonderfully done! However, this question comes from a precalculus textbook, which does not cover differentiation.