A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. The goal is to find the area of the largest possible Norman window with a perimeter of 35 feet?

A(max) = 72,13 ft²

Step-by-step explanation:

Total area of the Norman window is area of the semicircle plus area of the rectangle

Let call x radius of the semicircle, then the width of the rectangle is 2*x

Area of semicircle      A(sc) = π*r²/2    ⇒ A(sc) =  π*x²/2

Area of rectangle (  sides  2*x and y)  A(r)  = 2*x*y

Area of window  A(w)  =   π*x²/2  + 2* x*y

Perimeter of the window is:

Three sides of the rectangle   P(r)  =  2*y + 2*x  

Length of the semicircle  P(sc)  = π*x

Then  35  =  2*y + 2*x  +  π*x   ⇒  2*y  =  35 - 2*x - π*x  

y = 35 - x* ( 2  + π ) /2

Area of window as a function of x

A(x) =  π*x²/2  +2* [ 35 -x ( 2 +π ) /2 ]*x

A(x) =  π*x²/2  + x* (35 - 2*x - π*x )  

A(x) =  π*x²/2 + 35*x - 2*x² - π*x²

Taking derivatives on both sides of the equation

A´(x) = π*x + 35 - 4*x - 2* π*x

A´(x) = 0    -π*x  - 4*x  =  - 35 ⇒  - x ( 4 + π ) = -35  ⇒ x  = 4,90 ft

And  y = ( 35 -  x  - π*x )/2     ⇒ y =  (35 - 4,90 - 15,38 )/2

y = 7,36 ft

A(max) = 2*x*y * +  π*x²/2

A(max) = 72,13 ft²