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 31 feet

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To find the largest possible area of the Norman window, we need to optimize the shape while satisfying the given perimeter constraint.

Let's start by assigning variables to the width of the rectangle (w), the radius of the semicircle (r), and the length of the rectangle (l). Since the diameter of the semicircle is equal to the width of the rectangle, we have r = w/2.

The perimeter (P) of the Norman window can be calculated by summing the lengths of all sides:

P = 2w + πr + 2l

We are given that the perimeter is 31 feet, so we can write the following equation:

31 = 2w + π(w/2) + 2l

Simplifying the equation, we get:

31 = 2w + (πw/2) + 2l
31 = w(2 + π/2) + 2l
31 - 2l = w(2 + π/2)
(31 - 2l)/(2 + π/2) = w

Now, we have w expressed in terms of l. To find the maximum area, we need to consider that the area (A) of the Norman window is given by:

A = w*l + (πr^2)/2

Substituting the value of r in terms of w, we get:

A = w*l + (π(w/2)^2)/2
A = w*l + (πw^2)/8

To maximize the area, we need to express A solely in terms of l. Substituting the expression for w obtained earlier, we have:

A = [(31 - 2l)/(2 + π/2)]*l + (π[(31 - 2l)/(2 + π/2)]^2)/8

Now, we have A as a function of l. To maximize A, we can take the derivative of A with respect to l, set it equal to zero, and solve for l:

dA/dl = [(31 - 2l)/(2 + π/2)] + [(π(31 - 2l))/(8(2 + π/2))] = 0

Solving this equation, we can find the value of l that maximizes the area. Once we have l, we can substitute it back into the equation for w to find w. Finally, we can calculate the area A using the values of l and w obtained.

By directly solving for l and w, the resulting values may not be nice integers. However, the approach explained above provides a method to find the largest possible area given the perimeter constraint.