Maximizing Area of Norman Window

[macbowes]

Homework Statement

A Norman window has the shape of a rectangle surmounted by a semicircle of diameter equal to the width of the rectangle. If the perimeter of the window is 20 feet, what dimensions will admit the most light (maximize the area)?

[PLAIN]http://img163.imageshack.us/img163/5514/normanwindow.jpg

I drew that wonderful (I'm clearly not an artist, haha) in paint to help illustrate the problem.

x = 2r = w
y = l

Homework Equations

Perimeter = (2l + w) + \frac{2\pir}{2}

Area = (l * w) + \frac{\pir²}{2}

The Attempt at a Solution

Here's what I've done so far, I'll try and explain each step as I go.

#1. p = 2l + w + \frac{\piw}{2}= 20

I substituted w in for 2r order to reduce the amount of variables in my equation.

#2. A = (l * w) + \frac{\pi\frac{1}{4}w}{2}

I again substituted w in for 2r in order to reduce the amount of variables, this time for the area equation.

#3. 20 = 2l + (\frac{\piw + 2w}{2}

I'm now going to try and solve for l in the perimeter function. I made w \frac{2w}{2} so I can combine it with \frac{\piw}{2}. I want to leave \pi in that form instead of converting it to a decimal, but because of this I have to write it as \piw + 2w.

#4. 2l + (\frac{\piw + 2w}{2} - 20 = 0

Moved the 20 over to make it equal zero.

#5. -(\frac{\piw + 2w}{2} + 20 = 2l

Moved the 2l over and changed the signs of all the terms.

#6. -(piw + 2w) + 20 = 4l

Multiplied both sides by 2 to get rid of the fraction.

#7. \frac{-(\piw+2w)+40}{4} = l

Divided both sides by 4 to isolate l. Now that I know l, I can substitute the l into A(p) so that I can find the area as a function of w.

#8. A(p) = ((\frac{(\piw+2w)+40}{4})*w) + \frac{\pi\frac{1}{4}w}{2}

This is the function that resulted. I put this function into my calculator and graphed it. I then traced x = 20 to find the maximum area when p = 20 (the answer I got was 722.01).

This was as far as I got, but I suppose to find the dimensions of the window I could do A(20) to get w and then find my length by putting that into my perimeter function.

Honestly, I have not even the slightest clue if what I am doing was the right way to go about this problem because it seemed to just get more and more convoluted. Any guidance and/or assisstance would be greatly appreciated.

PS. I've never used the latex code on forums before, so hopefully I got it right >.<

EDIT: Wow, it's nothing like how I wanted to show it :(. I'm not really sure how to use the latex code properly, I thought I had everything right :(.

If you don't mind just quoting me so it will show you the latex code in my post in correcting it that would be awesome! Thanks.

 

[HallsofIvy]

Yes, w= 2r so the area of the rectangle is wh= 2rh and the area of the semi-circle is \pi r^2/2. The area of the Norman window is A= 2rh+ \pi r^2/ 2.

The perimeter includes the three sides of the rectangle, h+ h+ w= 2h+ 2r, together with half the circumference of a circle of radius r, 2\pi r/2= \pi r.
Your condition that the perimeter be 20 ft, then, is 2h+ 2r+ \pi r= 20 which we can write as h= 20- r- \pi r/2.

Putting that into the formula for area, A= 2rh+\pi r^2/ 2=40r- 2r^2- \pir^2+ \pi r^2/2=40r- (2- \pi)r^2. Differentiate that to find the r that gives maximum area.