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?
I got h=10.180619625,r=2.458141125 and Area = 59.54227.
Can someone check my work, i'm not v ery confident in my answer.
I didn't get that, I would recheck.
always loved that particular question.
I got a radius of 4.62 and a max area of 76.24
Oh ok i see where i got it wrong. ty for your help.
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To find the area of the largest possible Norman window with a given perimeter of 33 feet, we can use calculus optimization.
Let's assume the width of the rectangle is x. Since the diameter of the semicircle is equal to the width of the rectangle, the radius of the semicircle is x/2.
The perimeter of the Norman window can be expressed as:
Perimeter = 2 * (Length of rectangle) + (Circumference of semicircle)
33 = 2 * x + π * (x/2)
33 = 2x + (π/2)x
33 = (2 + π/2)x
Solving the equation for x, we get:
x = 33 / (2 + π/2)
x ≈ 10.18
So the width of the rectangle is approximately 10.18 feet.
Since the diameter of the semicircle is equal to the width of the rectangle, the radius of the semicircle is x/2 ≈ 5.09 feet.
To find the area of the window, we can sum the area of the rectangle and the area of the semicircle:
Area = (Area of rectangle) + (Area of semicircle)
Area = x * (x/2) + (π/2) * (x/2)^2
Area = (x^2/2) + (π/8)x^2
Area ≈ (10.18^2/2) + (π/8)(10.18^2)
Area ≈ 51.43 + 8.80
Area ≈ 60.23 square feet
Therefore, the area of the largest possible Norman window with a perimeter of 33 feet is approximately 60.23 square feet.
It seems there was a mistake in your calculations. The correct area is 60.23 square feet, not 59.54 square feet.
To find the maximum area of a Norman window with a given perimeter, we need to use optimization techniques. Here's how you can solve it step-by-step:
1. Let's denote the width of the rectangle as "w" and the radius of the semicircle as "r." Since the diameter of the semicircle is equal to the width of the rectangle, the radius is half the width, so r = w/2.
2. The perimeter of the window is given as 33 feet. The perimeter formula for the Norman window is P = 2w + πr + w, where 2w represents the two vertical sides of the rectangle, w represents the bottom side of the rectangle, and πr represents the curved top of the semicircle. Substituting the values, we have 33 = 2w + π(w/2) + w.
3. Simplify the equation: 33 = 2w + (πw/2) + w.
Combine like terms: 33 = (5π/2)w.
Divide both sides by (5π/2): w = (33 * 2) / (5π).
4. Now that we have the value of w, we can substitute it back into the equation for the radius of the semicircle: r = w/2. Calculate r using the value of w obtained in the previous step.
5. Once you have the values of w and r, you can calculate the area of the Norman window by finding the area of the rectangle (A_rec) and the area of the semicircle (A_sem). The area of the rectangle is A_rec = w * r, and the area of the semicircle is A_sem = (π * r^2) / 2. Add these two areas together to get the total area of the Norman window.
Double-check your work by comparing it with the results you provided:
- h = 10.180619625: This doesn't correspond to any of the measurements needed for the Norman window.
- r = 2.458141125: This value seems plausible for the radius of the semicircle.
- Area = 59.54227: This value doesn't match the area calculated using the given radius and width.
After rechecking your work, you found that you made an error. The correct radius is 4.62, not 2.458141125, and the maximum area is 76.24, not 59.54227.