# Maximizing Area of Norman Window

• macbowes
A'= 40- 2(2- \pi)r= 40- 4r+ 2\pir= 0 when 4r= 2\pir so that r= \pi/2. h= 20- \pi/2- \pi^2/4= (80- 2\pi- \pi^2)/4= (40- \pi- \pi^2/2)/2= (40- \pi- \pi^2)/4.
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.

Last edited by a moderator:
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.

## What is the concept behind the "Maximizing Area of Norman Window" problem?

The "Maximizing Area of Norman Window" problem is a mathematical optimization problem that involves finding the dimensions of a window with a fixed perimeter that maximizes the area of the window. It is named after the mathematician John Norman who first published the problem in 1958.

## What are the known dimensions and constraints in the "Maximizing Area of Norman Window" problem?

The known dimensions in the "Maximizing Area of Norman Window" problem are the perimeter of the window, which is fixed, and the shape of the window, which is a rectangle with a semicircle on top. The constraints are that the window must have a fixed perimeter and must fit within a given space, such as a wall or frame.

## How is the solution to the "Maximizing Area of Norman Window" problem determined?

The solution to the "Maximizing Area of Norman Window" problem is determined using calculus and optimization techniques. By finding the derivative of the area function with respect to the window's height or width, and setting it to zero, the critical points can be found. The critical points are then evaluated to determine which one gives the maximum area. In some cases, the solution may also require using the boundary conditions to find the optimal dimensions.

## Are there any real-world applications for the "Maximizing Area of Norman Window" problem?

Yes, the "Maximizing Area of Norman Window" problem has real-world applications in architecture and design. It can help architects and designers determine the optimal dimensions for windows in a building to maximize natural light and energy efficiency. It can also be used in manufacturing to optimize the use of materials and reduce costs.

## What are some variations of the "Maximizing Area of Norman Window" problem?

Some variations of the "Maximizing Area of Norman Window" problem include finding the dimensions of a window with a fixed perimeter that minimizes the area, or finding the dimensions of a window with a fixed area that maximizes the perimeter. There are also variations that involve different shapes, such as a square with a semicircle on top, or finding the dimensions of a window with a fixed perimeter that minimizes the amount of material used.

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