A Norman window is constructed by adjoining a semicircle to the top of a rectangular window . (The diameter of the semicircle is the same as the width of the rectangular) If the perimeter of the Norman window is 20 ft, find the dimensions that will allow the window to admit the most light.
To find the dimensions that will allow the window to admit the most light, we need to maximize the area of the window.
Let's denote the width of the rectangular part of the window as 'w' and the height as 'h'.
First, let's find the expression for the perimeter of the Norman window:
Perimeter = Length of the rectangular part + Circumference of the semicircle
The length of the rectangular part is equal to the sum of the width and height: 2w + h.
The circumference of the semicircle is half the circumference of a full circle, i.e., π(diameter) / 2.
Since the diameter of the semicircle is the same as the width of the rectangular part, the circumference of the semicircle is πw.
Now we can express the perimeter equation:
20 = 2w + h + πw
To find the dimensions that will admit the most light, we need to maximize the area. The area of the window is the sum of the area of the rectangular part and the area of the semicircle.
Area of the rectangular part = width * height = w * h
Area of the semicircle = 0.5 * π * radius^2 = 0.5 * π * (w/2)^2 = 0.25 * π * w^2
So our objective is to maximize the total area:
Total Area = Area of the rectangular part + Area of the semicircle = w * h + 0.25 * π * w^2
To maximize the total area, we can take the derivative of the area with respect to 'w' and set it equal to zero:
d/dw (w * h + 0.25 * π * w^2) = 0
Simplifying and solving for 'h':
h = -0.5πw + 10
Now we can substitute this value of 'h' back into the perimeter equation:
20 = 2w + (-0.5πw + 10) + πw
Simplifying the equation:
20 = 2w - 0.5πw + πw + 10
20 = (2 - 0.5π + π)w + 10
10 = (2.5 - 0.5π)w
Solving for 'w':
w = 10 / (2.5 - 0.5π)
Substituting this value of 'w' back into the expression for 'h':
h = -0.5π(10 / (2.5 - 0.5π)) + 10
Therefore, the dimensions that will allow the window to admit the most light are:
Width (w) = 10 / (2.5 - 0.5π) ft
Height (h) = -0.5π(10 / (2.5 - 0.5π)) + 10 ft
To find the dimensions that will allow the window to admit the most light, we need to determine the dimensions of the rectangular part of the window. Let's call the width of the rectangular part "w" and the height of the rectangular part "h".
To start, let's find the perimeter of the window. The perimeter is the sum of the lengths of all the sides.
Since the Norman window consists of a rectangular part and a semicircle, the perimeter is given by:
Perimeter = Length of rectangular + Circumference of semicircle
The length of the rectangular part is the same as the height, so it is "h". The circumference of the semicircle is half the circumference of a full circle, which is (π * diameter). The diameter is the same as the width, so it is "w".
Therefore, the perimeter can be expressed as:
Perimeter = h + (π * w)
Given in the problem, the perimeter is 20 ft. So we can write the equation:
20 = h + (π * w)
To find the dimensions that will allow the window to admit the most light, we need to maximize the area of the window.
The area of the window is the sum of the area of the rectangular part and half the area of the semicircle.
Area = Area of rectangular part + (1/2) * Area of semicircle
The area of the rectangular part is simply the product of its width and height:
Area of rectangular part = w * h
The area of the semicircle is half the area of a full circle, which is (1/2) * (π * r^2). The radius "r" is half the diameter "w/2". So, the area of the semicircle is:
Area of semicircle = (1/2) * (π * (w/2)^2)
Adding the area of the rectangular part and half the area of the semicircle, the total area is:
Area = w * h + (1/2) * (π * (w/2)^2)
Now, we need to find the dimensions that maximize the area. To do this, we can differentiate the area expression with respect to w and h separately, and set the derivatives equal to zero.
Taking the derivative of the area with respect to w:
d(Area)/dw = h + (π/2) * (w/2)
Setting this derivative equal to zero and solving for w:
h + (π/2) * (w/2) = 0
h + (πw/4) = 0
h = - (πw/4)
Similarly, taking the derivative of the area with respect to h:
d(Area)/dh = w
Setting this derivative equal to zero and solving for h:
w = 0
From these equations, we can see that we have found the dimensions that maximize the area of the Norman window. However, we need to ensure that these dimensions also satisfy the perimeter equation.
Substituting the values of w and h into the perimeter equation:
20 = h + (π * w)
20 = - (πw/4) + (π * w)
Simplifying:
20 + (πw/4) = πw
Multiplying both sides by 4/π:
(80/π) + w = 4w
Subtracting w from both sides:
(80/π) = 3w
Simplifying:
w = (80/3π)
Now that we have the value of w, we can substitute it back into the perimeter equation to find h:
20 = h + (π * (80/3π))
20 = h + (80/3)
Subtracting (80/3) from both sides:
20 - (80/3) = h
Finding a common denominator:
60/3 - 80/3 = h
-20/3 = h
So, the dimensions that will allow the window to admit the most light are:
Width (w) = (80/3π) ft
Height (h) = -20/3 ft
Please note that the negative value for h is not physically meaningful in this context, so we disregard it. The actual height of the rectangular part should be positive. Therefore, the dimensions that will allow the window to admit the most light are:
Width (w) ≈ 8.52 ft
Height (h) ≈ 6.67 ft
base = 2 r
height of rectangle = h
area is A
A = 2 r h + (1/2) pi r^2
perimeter = 2 r + 2 h + (1/2)(2 pi r)=20
(2 + pi)r + 2 h = 20
(1 + pi/2)r + h = 10
so
h = 10 - 2.57 r
A = 2 r (10-2.57r) + 1.57 r^2
A = 20 r - 5.14 r^2 + 1.57 r^2
or
A = 20 r - 3.57 r^2
dA/dr = 0 for max = 20 - 7.14 r
or
r = 2.8
all yours now