A Norman window has the shape of a rectangle surmounted by a semicircle. Find the dimensions of a Norman window of perimeter 39 ft that will admit the greatest possible amount of light. (Round your answers to two decimal places.)
To find the dimensions of a Norman window that will admit the greatest possible amount of light, we need to optimize the area of the window.
Let's start by visualizing the problem. A Norman window consists of a rectangle surmounted by a semicircle. The perimeter of the window is given as 39 ft.
Let's say the width of the rectangle is denoted by 'w' and the radius of the semicircle is denoted by 'r'. We can set up the equation for the perimeter of the window:
Perimeter = 2w + πr + w
39 = 2w + πr + w
Since we are trying to maximize the amount of light, the area of the window is our objective. The area of the window is the sum of the area of the rectangle and the area of the semicircle:
Area = lw + πr²/2
To find the maximum area, we can express the area in terms of a single variable and then find its maximum using calculus.
From the perimeter equation, we can solve for 'r' in terms of 'w':
2w + πr + w = 39
3w + πr = 39
r = (39 - 3w)/π
Now we can substitute this expression for 'r' in the area equation:
Area = lw + π((39 - 3w)/π)²/2
= lw + (39 - 3w)²/2π
To maximize the area, we can differentiate the area equation with respect to 'w', set it to zero, and solve for 'w':
d(Area)/dw = l - (3w(39 - 3w))/π
0 = l - (3w(39 - 3w))/π
To solve for 'w', we can simplify the equation:
3w(39 - 3w) = lπ
117w - 9w² = lπ
Now, we have a quadratic equation. To solve for 'w', let's set it equal to zero and solve for 'w':
9w² - 117w + lπ = 0
We can then use the quadratic formula:
w = (-b ± √(b² - 4ac)) / 2a
where a = 9, b = -117, and c = lπ.
However, we need more information to proceed with the calculation. The length 'l' of the window is not specified in the question. Please provide the length value so that we can calculate the dimensions of the Norman window that admits the greatest possible amount of light.
rectangle height h, width 2r
39 = 2r + 2h + pi r = r(2+pi) +2 h
so
2h = 39 - r(2+pi)
A = area = (1/2) pi r^2+ 2 r h
A = .5 pi r^2 + r[39 -r(2+pi) ]
A = .5 pi r^2 + 39 r - 2r^2 -pi r
A = (.5 pi -2) r^2 +(39-pi)r
dA/dr = 0 for max = 2(.5pi-2) r + (39-pi)
so
r = (39-pi)/(4-pi) etc