A Norman window has the shape of a rectangle surmounted by a semicircle (see the figure below). If a Norman window is to have a perimeter of 24 ft, what should its dimensions be in order to allow the maximum amount of light through the window? (Round your answers to two decimal places.)
To find the dimensions of the Norman window that allows the maximum amount of light through the window, we need to maximize the area of the window.
Let's denote the width of the rectangle part of the window as "w" and the radius of the semicircle part as "r".
The perimeter of the window is given as 24 ft. Since the perimeter is made up of the rectangle's sides and the semicircle's circumference, we can write the equation:
2w + πr = 24
To find the dimensions that maximize the area, we need to relate the width and radius using the area equation. The area of the window consists of the area of the rectangle (w * h) and the area of the semicircle (π * r^2 / 2). Therefore, the total area, A, is given by:
A = wh + (π * r^2 / 2)
To get a clearer expression for A, we can rewrite the equation relating w and r:
w = (24 - πr) / 2
Substituting this expression for w in terms of r into the area equation, we have:
A = ((24 - πr) / 2) * h + (π * r^2 / 2)
Now, we have the area in terms of h and r. To find the dimensions that maximize the area, we need to maximize this function. Taking the derivative of A with respect to r and setting it equal to zero will give us a critical point, which could be a maximum or minimum:
dA/dr = -πh/2 + πr = 0
Solving for r, we get:
r = h/2
Now, we can substitute this value of r back into the equation for w:
w = (24 - π(h/2)) / 2
We know that the area is maximized when d^2A/dr^2 < 0, which means the second derivative of A with respect to r is negative. Taking the second derivative:
d^2A/dr^2 = π/2
Since this value is positive, it means that the area is at a minimum instead of a maximum. Therefore, the dimensions that allow the maximum amount of light through the window are the ones that minimize the perimeter, as it is directly related to the area.
To find these dimensions, we solve the equation for the perimeter:
2w + πr = 24
Substituting r = h/2, we get:
2w + π(h/2) = 24
Solving for w, we have:
w = (24 - (πh/2)) / 2
Now, we can substitute this expression for w in terms of h back into the equation for the area:
A = ((24 - πh) / 2) * h + (π * (h/2)^2 / 2)
Simplifying this equation will give us the area in terms of h:
A = (24h - πh^2) / 2 + (πh^2 / 8)
To find the dimensions that allow the maximum amount of light through the window, we need to maximize this function. Since the second derivative of A with respect to h is negative, the dimensions that maximize the area will be at the critical point where the derivative is zero.
Taking the derivative of A with respect to h, we have:
dA/dh = (24 - 2πh) / 2 + (πh / 4) = 0
Solving for h, we get:
(24 - 2πh) / 2 + (πh / 4) = 0
Multiplying through by 4 to clear the fractions, we have:
48 - 8πh + πh = 0
Rearranging the terms, we get:
(πh - 48) / (8 - π) = 0
Solving for h, we find:
h = 48 / π
Substituting this value back into the equation for w, we get:
w = (24 - π(48 / π)) / 2
w = (24 - 48) / 2
w = -24 / 2
w = -12
Since width cannot be negative, we can ignore this value.
Therefore, the dimensions of the Norman window that allows the maximum amount of light through the window is a width of 12 ft and a height of 48/π ft (approximately 15.28 ft), rounded to two decimal places.