A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus t

Question

A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus the diameter of the semicircle is equal to the width of the rectangle. See the figure below.) If the perimeter of the window is 24 ft, find the value of x so that the greatest possible amount of light is admitted. (The picture shows a simple window that could be imagined by picturing a rectangle window that's longer lengthwise (vertically) and at the top, attached is a semicircle and the diameter of the semicircle matches the width of the rectangle piece but they're seemlessly conected. (optimization problem).

Explanation / Answer

When it says "so that the greatest possible amount of light is admitted" it means find the maximum possible area.

Let x be the width, which is equal to the diameter of the semi-circle.
Then the perimeter around just the semi-circle is (1/2) pi x.
Let y the height of the rectangular portion of the window.
Perimeter around just the rectangular portion of the window is x + 2y.
The total perimeter is (1/2) pi x + x + 2y = 24
Solve this equation for y:
2y = 24 - (1/2) pi x - x
y = 12 - (1/4) pi x - x/2
Then the area of the rectangular portion is xy.
The area of the semi-circle is (1/2) pi (x/2)^2.
The total area = A = (1/2) pi (x/2)^2 + xy
Substitute the expression for y found above into this last equation:
A = (1/2) pi (x/2)^2 + x(12 - (1/4) pi x - x/2 )
Simplify and combine like terms:
A = x^2(-pi - 4)/8 + 12x
Take the derivative and set it to zero:
A' = (1/4) (-4-pi)x + 12 = 0
Solve for x:
(1/4) (-4-pi)x = -12
Multiply by -4:
(4+pi)x = 48
Divide:
x = 48 / (4+pi) 6.7212 ft
y = 12 - (1/4) pi x - x/2 = 30 / (4+pi) 3.3606 ft

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