SOLUTION: Suppose there is a semicircle below the rectangle as well as above it. Find the dimensions of the window which admit most sunlight for a given perimeter 8m
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Question 70941: Suppose there is a semicircle below the rectangle as well as above it. Find the dimensions of the window which admit most sunlight for a given perimeter 8m
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Suppose there is a semicircle below the rectangle as well as above it. Find the dimensions of the window which admit most sunlight for a given perimeter 8m.
:
The most sunlight would be admitted at the max area and I seem to remember that
a circle is the shape to have max area for a given perimeter.
:
Let x = the radius of the two semi-circles
Then 2x = the length of the rectangle (horizontal measurement)
Width (W) will be the vertical measurement
:
Perimeter:
Twice the rectangle width + Circumference = 8
2W + 2pi*x = 8
:
2W = 8 - 2pi*x
:
W = (4 - pi*x); divided equation by 2
:
Area = Rectangle area + area of the circle
A = W(2x) + pi*x^2
:
A = 2x(4-pi*x) + pi*x^2; Substituted (4-pi*x) for W
A = 8x - 2pi*x^2 + pi*x^2
A = 8x - pi*x^2
:
Write it as a quadratic equation; y = area
y = -(pi*x^2) + 8x
:
Find the axis of symmetry to find the max area: a = -pi, b = 8
x = -8/2(-pi)
x = 1.27 meters is the radius of the semi circles at max area
:
Width of the rectangle = 4 - pi*x
W = 4 - pi*1.27
W = 4 - 4
W = 0:
Just as we suspected, the max area will be when there is no rectangle,
and the window is a circle.
:
Area if it is a circle: pi*1.27^2 = 5.067 sq meters
:
:
Let's say the width is 1 meter. Then the total circumference of semi-circles
will be 6 meters"
Find the radius if this is the case:
2*pi*r = 6
r = 6/(2pi)
r = .955 meters
:
The length of the rectangle will be twice that: 1.91 meters
Area of the rectangle 1*1.91 = 1.91 sq meters
:
The area of the circle = pi*.955^2 = 2.865 sq meters
Total area 1.91 + 2.865 = 4.77 sq meters, less than a circle will be.
:
Note to student, let me know if I am wrong about this or any comment you have
about this, I would glad to hear.
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