A Norman 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.)

Hope i hepled you on this question.

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