Question

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

Answer 1

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 !

Answer 2

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.

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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