A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus, the diameter of the semicircle is equal to the width of the rectangle.) If the perimeter of the window is 30 ft, find the dimensions of the window so that the greatest possible amount of light is admitted.
I found the base length that is equal to 60/(4+pi) ft but I can't find the total height. Thank you!!
One of the popular questions in Calculus texts.
Let the radius of the circle be x
then the width of the rectangle = 2x
let the height of the rectangle be y
2x + 2y + halfcircle = 30
2x + 2y + (1/2)(2πx) = 30
2x + 2y + πx = 30
2y = 30 - 2x - πx
y = 15 - x - πx/2
area of window = A
= 2xy + (1/2)πx^2
= 2x(15 - x - (1/2)πx) + (1/2)πx^2
= 30x - 2x^2 - πx^2 + (1/2)πx^2
dA/dx = 30 - 4x - 2πx + πx
= 0 for a max of A
30 = 4x+πx
30 = x(4+π)
x = 30/(4+π) = appr 4.2
base = 2x = 60/(4+π) ---> good for you, you had that
now y = 15 - 4.2 - π(4.2)/2 = appr 4.2 , ahhh, the same
So the rectangle is
60/(4+π) by 30/(4+π)
or
appr 8.4 at the base and 4.2 high
To find the total height of the Norman window, we need to break down the problem into two parts: the rectangle and the semicircle.
Let's start with the rectangle.
The perimeter of a rectangle is given by the formula: 2(length + width).
Let's say the length of the rectangle is L, and the width is W.
So, for the rectangle part of the Norman window, we have:
2(L + W) = Perimeter of rectangle = 30 ft
Now, let's move on to the semicircle.
The circumference of a semicircle is given by the formula: pi(radius) + 2(radius).
Since the diameter of the semicircle is equal to the width of the rectangle, the radius will be half of the width. So, the radius will be W/2.
Therefore, for the semicircle part of the Norman window, we have:
(pi * (W/2)) + 2(W/2) = Perimeter of semicircle = Perimeter of rectangle = 30 ft
Now, let's solve the two equations together to find the dimensions of the window.
2(L + W) = 30 ---(equation 1)
(pi * (W/2)) + 2(W/2) = 30 ---(equation 2)
From equation 2, we can simplify it to:
(pi * W/2) + W = 30
(pi * W + 2W)/2 = 30
(pi * W + 2W) = 60
W(pi + 2) = 60
W = 60/(pi + 2)
Now, let's substitute the value of W in equation 1 to find L:
2(L + (60/(pi + 2))) = 30
2L + (120/(pi + 2)) = 30
2L = 30 - (120/(pi + 2))
2L = (30(pi + 2) - 120)/(pi + 2)
L = (30(pi + 2) - 120)/(2(pi + 2))
So, the dimensions of the window are:
Length (L) = (30(pi + 2) - 120)/(2(pi + 2)) ft
Width (W) = 60/(pi + 2) ft
This should give you the total height and other dimensions of the Norman window.
To find the dimensions of the window so that the greatest possible amount of light is admitted, we need to find the total height of the window.
Let's denote the width of the rectangle as "w" and the height of the rectangle as "h". Since the diameter of the semicircle is equal to the width of the rectangle, we can also write the width of the semicircle as "w".
The perimeter of the window is given as 30 ft, so we can write the equation:
Perimeter = 2w + h + πw = 30
Now, let's substitute the expression for the base length that you found into the equation:
2(60/(4+π)) + h + π(60/(4+π)) = 30
Now, we can solve this equation for the height "h".
1. Simplify the expression:
120/(4+π) + h + 60π/(4+π) = 30
2. Combine like terms:
120 + h(4+π) + 60π = 30(4+π)
3. Distribute and simplify:
120 + 4h + hπ + 60π = 120 + 30π
4. Combine like terms:
4h + hπ + 60π = 30π
5. Simplify:
4h + hπ + 60π - 30π = 0
6. Combine like terms:
4h + (h + 30)π = 0
7. Factor out "h":
h(4 + π) + 30π = 0
Now, we can solve this equation for "h".