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

Answer:

132.233 ft2

Step-by-step 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:

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