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Norman window

by Andrew
(New York)











































A Norman window has the shape of a rectangle surmounted by a semicircle, as shown in the figure below. If the perimeter of the window is 30 feet, find the dimensions of the window, to the nearest hundredth, that admits the greatest possible amount of light.

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Nov 06, 2011
Maximize Area of Norman Window
by: Staff

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Part II


Calculate the value of r which will provide the maximum area for the window

Differentiate the equation for A with respect to r

A = (-2 - (π/2))r² + 30r


0 = 2*[-2 - (π/2)]r + 30

Solve for r

0 = 2*[-2 - (π/2)]r + 30

2*[-2 - (π/2)]r + 30 = 0

(-4 - π)r + 30 = 0

(-4 - π)r + 30 - 30 = 0 - 30

(-4 - π)r + 0 = 0 - 30

(-4 - π)r = 0 - 30

(-4 - π)r = - 30

(-4 - π)r / (-4 - π) = - 30 / (-4 - π)

r * (-4 - π) / (4- - π) = - 30 / (-4 - π)

r * (1) = - 30 / (-4 - π)

r = -30 / (-4 - π)

r = 4.2007436513367 ft

the final answer to your question:

light through the window will be maximized when r = 4.20 ft



Open the following link to see a graph of the equation for the area of the window as a function of r


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http://www.solving-math-problems.com/images/graph-area-vs-radius-2011-11-05.png




The only other dimension to calculate is x

x = [30 - r(π + 2)]/2

substitute 4.20 ft for the value of r

x = [30 - 4.20*(π + 2)]/2

x = 4.20


the final dimensions for the Norman Window are


Open the following link to see the final dimensions for the Norman Window


(1) If your browser is Firefox, click the following link to VIEW ; or if your browser is Chrome, Internet Explorer, Opera, or Safari (2A) highlight and copy the link, then (2B) paste the link into your browser Address bar & press enter:

Use the Backspace key to return to this page:

http://www.solving-math-problems.com/images/graph-window-final-dimensions-2011-11-05.jpg



Thanks for writing.

Staff
www.solving-math-problems.com



Nov 06, 2011
Maximize Area of Norman Window
by: Staff


Question:

by Andrew
(New York)


A Norman window has the shape of a rectangle surmounted by a semicircle, as shown in the figure below. If the perimeter of the window is 30 feet, find the dimensions of the window, to the nearest hundredth, that admits the greatest possible amount of light.




Answer:



The answer:

Open the link shown below to view all the dimensions of the window:

(1) If your browser is Firefox, click the following link to VIEW ; or if your browser is Chrome, Internet Explorer, Opera, or Safari (2A) highlight and copy the link, then (2B) paste the link into your browser Address bar & press enter:

Use the Backspace key to return to this page:

http://www.solving-math-problems.com/images/window-dimensions-2011-09-22.jpg


This problem has two unknown values: r and x

It will require two equations


1st equation (based on the perimeter)

perimeter = 30 ft

πr + x + 2r + x = 30

solve for x

πr + x + 2r + x = 30

πr + 2r + x + x = 30

r(π + 2) + 2x = 30

r(π + 2) - r(π + 2) + 2x = 30 - r(π + 2)

0 + 2x = 30 - r(π + 2)

2x = 30 - r(π + 2)

2x/2 = [30 - r(π + 2)]/2

x*(2/2) = [30 - r(π + 2)]/2

x*(1) = [30 - r(π + 2)]/2

x = [30 - r(π + 2)]/2


2nd equation (based on the area)

Area of window = area of half circle + area of rectangle

A = (πr²)/2 + 2rx


Substitute [30 - r(π + 2)]/2 for x

A = (πr²)/2 + 2r*[30 - r(π + 2)]/2

A = (πr²)/2 + r*[30 - r(π + 2)]*(2/2)

A = (πr²)/2 + r*[30 - r(π + 2)]*(1)

A = (πr²)/2 + r*[30 - r(π + 2)]

A = (πr²)/2 + 30r - r²(π + 2)

A = (πr²)/2 - r²(π + 2) + 30r

A = r² * [(π/2) - (π + 2)] + 30r

A = r² * [-2 - (π/2)] + 30r

A = [-2 - (π/2)] r² + 30r


The equation for A (the area of the window) can be plotted and solved graphically, or it can be solved analytically.

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