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 33 feet?

Let x = the width of the rectangle.

The perimeter of the window is 33 feet, so:
2x + πx = 33

Solving for x, we get:
x = (33 - π)/2

The area of the window is the area of the rectangle plus the area of the semicircle:
A = x(x + 2πx/2)
= x^2 + πx^2/2
= (33 - π)^2/4 + (π/2)(33 - π)^2/4
= (33^2 - 2π33 + π^2)/4 + (π/2)(33^2 - 2π33 + π^2)/4
= (33^2 - 2π33 + π^2 + (π/2)(33^2 - 2π33 + π^2))/4
= (33^2 - 2π33 + (3π^2 + 33π)/2)/4
= (33^2 + 33π + 3π^2)/4
= (33 + 3π)(33 + π)/4
= 1089π/4

Therefore, the area of the largest possible Norman window with a perimeter of 33 feet is 1089π/4 square feet.

To find the area of the largest possible Norman window with a perimeter of 33 feet, we need to determine the dimensions of the window that would result in the maximum area.

Let's start by assigning variables to the dimensions of the rectangle. Let's say the width of the rectangle is 'w'. Since the diameter of the semicircle is equal to the width of the rectangle, the radius of the semicircle would be 'w/2'. The height of the rectangle would be 'h'.

The perimeter is the sum of all the sides:
Perimeter = 2 * (width + height) + circumference of semicircle

Since the circumference of a semicircle is half the circumference of a full circle, we can calculate it using the formula:

Circumference of semicircle = (π * diameter) / 2 = (π * w) / 2

Now we can express the perimeter equation as:
33 = 2 * (w + h) + (π * w) / 2

Simplifying the equation:
33 = 2w + 2h + (π * w) / 2
33 = 2(w + h) + (π * w) / 2

To proceed further, we need to know the value of π. Since it's not mentioned in the question, we'll use the approximation of π as 3.14 for calculation purposes.

33 = 2(w + h) + (3.14 * w) / 2

To find the dimensions that result in the maximum window area, we need to express the area in terms of a single variable. The area of the window can be calculated as the sum of the areas of the rectangle and the semicircle:

Area = rectangle area + semicircle area

Rectangle area = width * height = w * h
Semicircle area = (π * radius^2) / 2 = (π * (w/2)^2) / 2

Therefore, the total area becomes:
Area = w * h + (π * (w/2)^2) / 2

To maximize the area, we can express 'h' in terms of 'w' so that the total area is a function of a single variable:

h = (33 - 2w - (3.14 * w) / 2) / 2

Substituting this value of 'h' in the area equation, we get:
Area = w * [(33 - 2w - (3.14 * w) / 2) / 2] + (π * (w/2)^2) / 2

Now, we can find the derivative of the area function with respect to 'w' and set it equal to zero to find the maximum:

d(Area)/dw = 0

Differentiating the equation using product rule and simplifying:
(33 - 2w - (3.14 * w) / 2) / 2 - w/2 + (3.14 * w) / 8 = 0

Multiplying through by 8 to clear the denominators:
33 * 8 - 16w - 3.14w + 16w + 3.14w = 0
264 - 0.86w = 0
0.86w = 264
w = 264 / 0.86
w ≈ 306.98

Since the width of the rectangle cannot be negative, we take the positive solution. Now we can find the corresponding height:

h = (33 - 2w - (3.14 * w) / 2) / 2
h = (33 - 2(306.98) - (3.14 * 306.98) / 2) / 2
h ≈ 4.11

Now let's calculate the area using the obtained dimensions:
Area = w * h + (π * (w/2)^2) / 2
Area = 306.98 * 4.11 + (3.14 * (306.98/2)^2) / 2
Area ≈ 1252.36 square feet

Therefore, the largest possible area for the Norman window with a perimeter of 33 feet is approximately 1252.36 square feet.

To find the area of the largest possible Norman window with a perimeter of 33 feet, we need to optimize the shape's dimensions. Let's break it down step by step.

1. Let's assume the width of the rectangle is represented by 'w' feet. Since the diameter of the semicircle is equal to the width of the rectangle, the radius of the semicircle would be 'w/2' feet.

2. The perimeter of the Norman window is the sum of the lengths of all sides. In this case, it includes the length of the rectangle (2 times the width) and the semicircle's circumference.

Perimeter = 2w + π(w/2) = 33 feet

3. Simplify the equation by distributing π to (w/2):

2w + (π/2)w = 33

4. Solve for 'w' by isolating it on one side of the equation. Subtract 2w from both sides:

(π/2)w = 33 - 2w

5. To remove the fraction, multiply both sides by 2/π:

w = (66 - 4w)/π

6. Now, we have an expression for the width 'w' in terms of 'π'. We can use this expression to find the value of 'w'.

7. To maximize the area of the Norman window, we need to maximize the width 'w' since the height is equal to 'w'.

8. Rearranging the equation, we have:

w + (4w/π) = 66

(4w/π) = 66 - w

4w = (66 - w)π

4w = 66π - wπ

5w = 66π

w = (66π)/5

9. Now that we have the value of 'w', we can calculate the area of the Norman window.

Area = width * height
= w * w
= (66π/5) * (66π/5)
= (4356π²)/25
≈ 1770.67 square feet

Therefore, the area of the largest possible Norman window with a perimeter of 33 feet is approximately 1770.67 square feet.