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?
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Let the rectangle have width $$ x $$ and height $$ y $$. The semicircle on top has a diameter equal to $$ x $$ and hence a radius of $$\frac{x}{2}$$.

The total area of the Norman window is the sum of the rectangle’s area and the semicircle’s area:
$$
A = xy + \frac{1}{2}\pi\left(\frac{x}{2}\right)^2 = xy + \frac{\pi x^2}{8}.
$$

The perimeter of the window consists of the curved edge of the semicircle, the two vertical sides of the rectangle, and the bottom side of the rectangle. (The top side of the rectangle is removed, being replaced by the semicircle.) Thus,
$$
\text{Perimeter} = x + 2y + \frac{1}{2}(2\pi\cdot \frac{x}{2}) = x + 2y + \frac{\pi x}{2}.
$$
We are given that the total perimeter is 41 feet:
$$
x + 2y + \frac{\pi x}{2} = 41.
$$

Solve for $$ y $$ in terms of $$ x $$:
$$
2y = 41 - x - \frac{\pi x}{2}, \qquad \text{so} \qquad y = \frac{41}{2} - \frac{x}{2} - \frac{\pi x}{4}.
$$
It is convenient to combine the $$ x $$–terms:
$$
y = \frac{41}{2} - x\left(\frac{1}{2} + \frac{\pi}{4}\right) = \frac{41}{2} - x\left(\frac{2+\pi}{4}\right).
$$

Now substitute this expression for $$ y $$ into the area $$ A $$:
$$
A(x) = x\left(\frac{41}{2} - \frac{2+\pi}{4}\,x\right) + \frac{\pi x^2}{8}.
$$
Distribute the $$ x $$:
$$
A(x) = \frac{41}{2}x - \frac{2+\pi}{4}\,x^2 + \frac{\pi x^2}{8}.
$$
Write the quadratic terms with a common denominator:
$$
\frac{2+\pi}{4}\,x^2 = \frac{2(2+\pi)}{8}\,x^2,
$$
so that
$$
A(x) = \frac{41}{2}x - \frac{2(2+\pi)}{8}\,x^2 + \frac{\pi x^2}{8}.
$$
Combine the $$ x^2 $$ terms:
$$
A(x) = \frac{41}{2}x - \frac{1}{8}\Bigl[2(2+\pi)-\pi\Bigr]x^2.
$$
Simplify the bracket:
$$
2(2+\pi) - \pi = 4 + 2\pi - \pi = 4+\pi.
$$
Thus,
$$
A(x) = \frac{41}{2}x - \frac{4+\pi}{8}\,x^2.
$$

This quadratic function in $$ x $$ opens downward, so its maximum occurs where its derivative is zero. Differentiate with respect to $$ x $$:
$$
A'(x) = \frac{41}{2} - \frac{4+\pi}{4}\,x.
$$
Set $$ A'(x) = 0 $$ for a maximum:
$$
\frac{41}{2} - \frac{4+\pi}{4}\,x = 0 \quad \Longrightarrow \quad \frac{41}{2} = \frac{4+\pi}{4}\,x.
$$
Solve for $$ x $$:
$$
x = \frac{41}{2} \cdot \frac{4}{4+\pi} = \frac{82}{4+\pi}.
$$

Next, substitute $$ x $$ back into the expression for $$ y $$:
$$
y = \frac{41}{2} - \frac{2+\pi}{4}\left(\frac{82}{4+\pi}\right).
$$
Notice that
$$
\frac{82}{4} = \frac{41}{2},
$$
so
$$
y = \frac{41}{2} - \frac{41}{2}\cdot\frac{2+\pi}{4+\pi} = \frac{41}{2}\left(1 - \frac{2+\pi}{4+\pi}\right).
$$
Simplify the fraction:
$$
1 - \frac{2+\pi}{4+\pi} = \frac{(4+\pi) - (2+\pi)}{4+\pi} = \frac{2}{4+\pi}.
$$
Thus,
$$
y = \frac{41}{2} \cdot \frac{2}{4+\pi} = \frac{41}{4+\pi}.
$$

Now, substitute $$ x = \frac{82}{4+\pi} $$ into the area function:
$$
A_{\max} = \frac{41}{2}\left(\frac{82}{4+\pi}\right) - \frac{4+\pi}{8}\left(\frac{82}{4+\pi}\right)^2.
$$
First, write the terms clearly:
$$
A_{\max} = \frac{41 \cdot 82}{2(4+\pi)} - \frac{82^2}{8(4+\pi)}.
$$
Notice that $$ 41 \cdot 82 = 82 \cdot 41 $$ and we can combine the expression:
$$
A_{\max} = \frac{82}{8(4+\pi)}\Bigl[4\cdot41 - 82\Bigr].
$$
Since $$ 4\cdot41 = 164 $$ and $$ 164 - 82 = 82 $$, we obtain:
$$
A_{\max} = \frac{82 \cdot 82}{8(4+\pi)} = \frac{82^2}{8(4+\pi)}.
$$

Thus, the area of the largest possible Norman window with a perimeter of 41 feet is
$$
\boxed{\frac{82^2}{8(4+\pi)} \text{ square feet}}.
$$

A numerical approximation (using $$\pi \approx 3.14$$) gives:
$$
A_{\max} \approx \frac{6724}{8(7.14)} = \frac{6724}{57.12} \approx 117.7 \text{ square feet}.
$$
The largest possible area of the Norman window with a perimeter of 41 feet is $$ \frac{82^2}{8(4+\pi)} $$ square feet, which is approximately 117.7 square feet.

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?
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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?
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