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?

A = x(31/2-1/4 (2+pi) x)+.5pi(x/2)^2

This is what I'm stuck on:
Find A′ and maximize the area

I have no idea what I'm supposed to do after I find the derivative

h is height of rectangle1

31 = 2 h + D + pi D/2
2 h = 31 -D(1+pi/2)
h = 15.5 - (D/2) (1+pi/2)

area = D h + (1/2)(pi D^2/4)
A = D (15.5 -1.29 D) + (1/8)pi D^2
A = 15.5 D - .897 D^2
dA/dD = 0 at max = 15.5 - 1.79 D
D = 8.64 ft
then h = 4.4 ft
check my arithmetic !

Thank you for my help. Can you explain what to plug in to get the answer?

My calculus teacher told me this is one of the harder problems, and it wouldn't be on the final. Despite this, I'm still really lost on exactly what this question is asking. This is similar to related rates, and that is my weakest section, by far.

we had

A = 15.5 D - .897 D^2

and we know

D = 8.64

A = 15.5 (8.64) - .897(8.64^2)
= 67 ft^2

I plugged in 8.64 * 4.4+.5(pi8.64^(2/4)), I'm sure that's how you find the solution

Oh okay, I guess that's what I was doing wrong.

Hey, I went back and got the exact value, because the answer isn't coming out correct on the practice program.

I did

15.5(8.637013338)-.8973009183(8.637013338) and I got
126.1237067
and that is still incorrect.

I'm not sure what error I am making. I hope I can figure this out

Whoops I forgot to square the second part, but it still did not produce the correct answer I got 66.9.....

I do not know - ask teacher. Unless we made arithmetic errors, that should be right. Perhaps in terms of pi or something ?

Let x be the width and y be the height of the window. So the radius of the semicircle at the top is r=x/2.

The perimeter 31=x+2y+πr . So x+2y+πx/2=31.
Solve this equation for y.
The area A=xy+12πr^2=xy+1/2π(x/2)^2.
Eliminate y from A by substituting y by its value found from the perimeter.
Now A should be a function of x only. Find A′ and maximize the area
-----------------------------

That's the hint for the problem, that's how I got the derivative I posted, and I was lost after that point.

After finding the derivative of the area function with respect to x, denoted as A', you can set it equal to zero to find any critical points. Critical points are values of x where the slope of the area function is zero, which can potentially correspond to the maximum or minimum points of the function.

Once you find the critical points, you can check if they correspond to a maximum or minimum by checking the second derivative. If the second derivative is positive at a critical point, it indicates a minimum value, while if it is negative, it indicates a maximum value.

Let's go through the steps to find A' and determine the maximum area for the Norman window:

1. Start with the area function: A = x(31/2 - 1/4(2 + π)x) + 1/2π(x/2)^2.

2. Find the derivative of A with respect to x, denoted as A'. Since A involves both x and constants, the derivative will include product rule and chain rule:

A' = (31/2 - 1/4(2 + π)x) + x(-1/4(2 + π)) + 0.5π(1/2)(1/2)x(1/2-1) = 31/2 - 1/4(2 + π)x - 1/4(2 + π)x - πx^2/8.

Simplifying A': A' = 31/2 - 1/2(2 + π)x - πx^2/8.

3. Set A' equal to zero to find the critical points:

31/2 - 1/2(2 + π)x - πx^2/8 = 0.

4. Solve the equation to find the possible values of x.

5. Once you have the possible values of x, you need to determine which one corresponds to a maximum area. To do this, take the second derivative of the area function, A'', and substitute each critical point:

A'' = -1/2(2 + π) - πx/4.

6. Plug in each critical point value of x into A''. If the result is negative, it indicates a maximum area.

7. The critical point that corresponds to a negative value for A'' is the maximum area of the Norman window.

Once you find the x-value for the maximum area, substitute it back into the area function A to calculate the actual maximum area.