A Norman window is a window with a semi-circle on top of regular rectangular window. What should be the dimensions of the window to allow in as much light as possible, if there are only 12 ft of the frame material available?

Using the perimeter and area formular for a rectangle and semicircle, the dimension of the window should have a length of 1.68 ft and the width should also be 1.68 ft

Using the Parameters :

Perimeter of semicircle = πr + 2r

Hence, perimeter of window can be expressed thus :

Area of window :

Making 2L the subject of the equation :

2L = 12 - 2r - πr - - - (3)

Substitute 2L into (2)

A = r(12 - 2r - πr) + 0.5πr^2

A = 12r - 2r^2 - πr^2 + 0.5πr^2

A = 12r - 2r^2 - 0.5πr^2

Take the derivative of A to obtain maximum value of r

dA/dr = 12 - 4r - πr = 0

4r + πr = 12

r = 12 / ( 4 + π)

r = 12 / (4 + 3.14)

r = 1.68 ft.

Recall, width = 2r

width of the window = 2 * 1.68 = 3.36 ft.

From (3) :

2L = 12 - 2r - πr

12 - 2*1.68 - 1.68(3.14)

2L = 3.3648

L = 3.3648 / 2

L = 1.68.

Hence, the dimension of the window should be : 1.68 by 1.68

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Side lengths = 1.68 ft and width = 3.36 ft.

Step-by-step explanation:

Let the side lengths of the window be L and the width = 2r ( r is also the radius of the semi-circle).

So we have

Perimeter = 2L + 2r + πr = 12

Area = 2rL + 0.5πr^2

From the first equation

2L = 12 - 2r - πr

Substitute for 2L in the equation for the area:

A = r(12 - 2r -  πr) + 0.5πr^2

A = 12r - 2r^2 -  πr^2 + 0.5πr^2

A =  12r - 2r^2 - 0.5πr^2

We need to find  r for the maximum area:

Finding the derivative and equating to zero:

A' = 12 - 4r -  πr = 0=

4r + πr = 12

r = 12 / ( 4 +  π)

r = 1.68 ft.

So the  width of the window = 2 * 1.68 = 3.36 ft.

Now  2L = 12 - 2r - πr

=  12 - 2*1.68 - 1.68π

= 3.36

L = 1.68.

slope is 35 and 20 miles is the y-intercept.

step-by-step explanation: