Anorman window has the shape of a rectangle surmounted by a semicircle. find the dimensions of a norman window of perimeter 30 ft that will admit the greatest possible amount of light. (round your answers to two decimal places.)

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 = 30
Solve this equation for y:
2y = 30 - (1/2) pi x - x
y = 15 - (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(15 - (1/4) pi x - x/2 )
Simplify and combine like terms:
A = x^2(-pi - 4)/8 + 15x
Take the derivative and set it to zero:
A' = (1/4) (-4-pi)x + 15 = 0
Solve for x:
(1/4) (-4-pi)x = -15
Multiply by -4:
(4+pi)x = 60
Divide:
x = 60 / (4+pi) ≈ 8.4 ft
y = 15 - (1/4) pi x - x/2 = 30 / (4+pi) ≈ 4.2 ft