Need Help with Optimization: area of Norman window

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Kristally

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Nov 12, 2006
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I need help solving several problems, but I will post only one right now.

This is the problem:
The perimeter of the following Norman window is 12 meters. Find the dimensions of the window that gives the greatest area.
math1.jpg

(this is a simular image as on the paper that I created in photoshop... not to scale)

So far I have:
P=perimeter
P=c+2l+w
c=(pi)w
12=c+2l+w => (pi)w+2l+w => (pi + 1)w = 2l
l= (12 - (pi + 1))/2

After this I do not know what to do, or even if what I did was correctly done to solve the problem.
Can someone help me please?

[/b]
 
Re: Need Help with Optimization

Kristally said:
I need help solving several problems, but I will post only one right now.

This is the problem:
The perimeter of the following Norman window is 12 meters. Find the dimensions of the window that gives the greatest area.
math1.jpg

(this is a simular image as on the paper that I created in photoshop... not to scale)

So far I have:
P=perimeter
P=c+2l+w
c=(pi)w
12=c+2l+w => (pi)w+2l+w => (pi + 1)w = 2l
l= (12 - (pi + 1))/2

After this I do not know what to do, or even if what I did was correctly done to solve the problem.
Can someone help me please?

[/b]
Click to expand...

Your Perimiter is a bit off. If W is the width of the rectangle and L the length, then P = (Pi*W)/2 + 2L + W = ((2+Pi)/2)W + 2L.

Also, A = [Pi*(W/2)<sup>2</sup>]/2 + LW = (Pi*W<sup>2</sup>)/8 + LW.

So, \(\displaystyle \L 12 =\frac{2+ \pi}{2}W + 2L\)
\(\displaystyle \L \Rightarrow L = 6 - \frac{2+ \pi}{4}W\)

Plug our formula in for L in our Area equation.

So, \(\displaystyle \L A = \frac{\pi W^2}{8} + LW\)

\(\displaystyle \L \Rightarrow A = \frac{\pi W^2}{8} + (6 - \frac{2+ \pi}{4}W)W = ( \frac{\pi}{8} - \frac{2+ \pi}{4})W^2 + 6W\)

\(\displaystyle \L = \frac{- \pi - 4}{8}W^2 + 6W\).

Then...

\(\displaystyle \L \frac{dA}{dW} =\) what?

Then, you will need to find when A is a maximum using A' above. Can you take it from here?
 
Wow, thank you so much.

So c= (pi*w)/2 instead of just c=(pi*w)?

Then take the derivated of the last equation and it would be be (dA/dW)=(-3pi-8)w+12. Right?
 
Kristally said:
Wow, thank you so much.

So c= (pi*w)/2 instead of just c=(pi*w)?

Then take the derivated of the last equation and it would be be (dA/dW)=(-3pi-8)w+12. Right?
Click to expand...
I edited my post before you got a chance to reply because I made a mistake, but I can see you have the right idea. The reason C = pi*w/2 is because P of a circle is Pi*Diameter. You have Half a circle with Diameter W.
 
Alright I understand that.

Then the derivative of that would be ((-pi- 4)/ 4)*w + 6. Right?

And than set it to 0 to find out w? Or something else?
 
Kristally said:
Alright I understand that.

Then the derivative of that would be ((-pi- 4)/ 4)*w + 6. Right?

And than set it to 0 to find out w? Or something else?
Click to expand...
You got it!
 
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