a norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular window. find the dimensions of a norman window of maximum area if the total perimeter is 72 units. Base value is 21.
if the rectangle is x wide by y high, then the diameter of the window is x.
So
x + 2y + π/2 x = 72
thus, y = [72-(1 + π/2)x]/2
The area is
a = xy + π/8 x^2
= x[72-(1 + π/2)x]/2 + π/8 x^2
= 36x - (1+π/4)x^2 + π/8 x^2
= 36x - (1 + 3π/8)x^2
da/dx = 36 - (2 + 3π/4)x
da/dx=0 when
x = 36/(2+ 3π/4)) = 144/(8+3π)
now you can easily find y.
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right?
To find the dimensions of a Norman window of maximum area with a total perimeter of 72 units, we can use the calculus concept of optimization.
Let's denote the width of the rectangular part of the window as "x" units. Since the base value is given as 21, the height of the rectangle would be 21/x units.
The perimeter of the rectangular part would be 2x and the circumference of the semicircle would be πr, where r is the radius. Since the semicircle joins the top of the rectangular window, the radius would be half of the width, which is x/2.
Therefore, the total perimeter is given by:
Perimeter = 2x + π(x/2)
We know that the total perimeter is 72 units, so we can set up the equation:
72 = 2x + π(x/2)
To find the dimensions for maximum area, we need to maximize the area of the window. The area of the rectangular part would be x * (21/x) = 21 units^2. The area of the semicircle would be (π/2)(x/2)^2.
Therefore, the total area would be:
Area = 21 + (π/8)x^2
To find the maximum area, we can differentiate the area equation with respect to x and set it equal to zero.
d(Area)/dx = (π/4)x = 0
Solving for x:
x = 0
However, x = 0 is not a valid solution in this context since it would result in a degenerate window. We need to find the critical points by setting d(Area)/dx = (π/4)x = 0. In this case, x = 0 is a critical point, but it does not provide a maximum area.
To determine the maximum area, we need to check the endpoints of the possible range for x. In this case, since we are dealing with a real-world window, the width cannot be negative or zero, so x > 0.
Since the width of the rectangular part of the window cannot exceed the total perimeter divided by 2, we have x ≤ 72/2 = 36.
Therefore, the only possible values for x that we need to consider are within the interval 0 < x ≤ 36.
To find the dimensions of the Norman window with maximum area, we can evaluate the area equation for the critical points and endpoints:
When x = 0, Area = 21 (Degenerate window, not valid)
When x = 36, Area = 21 + (π/8)(36)^2 ≈ 274.89 units^2
Now we compare the areas at the critical points and endpoints to determine the maximum area:
Maximum Area: 274.89 units^2
Therefore, the dimensions of the Norman window with maximum area, given a total perimeter of 72 units and a base value of 21, are a width of 36 units for the rectangular part and a corresponding height of 21/36 = 0.583 units.