Question

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. What is the area of the largest possible Norman window with a perimeter of 41 feet?

Answer 1

132.233 ft2

Explanation:

Let's call the width of the rectangle 'w' and the length 'x'. So the area of the semicircle is:

`A_1 = \pi*radius^2/2`

`A_1 = \pi*(w/2)^2/2`

`A_1 = \pi/8*w^2`

And the area of the rectangle is:

`A_2 = w*x`

If the perimeter of the window is 41 feet, we have:

`Perimeter = length + 2*width + \pi*radius`

`41 = x + 2*w + \pi*w/2`

`x = 41 - w(2 + \pi/2)`

Now, the equation for the total area of the window is:

`A = A_1 + A_2 = \pi/8*w^2 + w*x`

`A = \pi/8*w^2 + w*(41 - w(2 + \pi/2))`

`A = (\pi/8-2 - \pi/2)*w^2 + 41w = -3.1781w^2 + 41w`

To find the maximum area, we can find the x-coordinate of the vertex of the quadratic equation:

`x\_vertex = -b / 2a = -41 / (-3.1781*2) = 6.45`

So the width that gives us the maximum area of the window is 6.45 feet, and the area will be:

`A = -3.1781w^2 + 41w = -3.1781*(6.45)^2 + 41*6.45 = 132.233\ ft^2`