Calculus optimization - A Norman window has the shape of a rectangle surmounted by a semicircle?

Question

A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus, the diameter of the semicircle is equal to the width of the rectangle.). If the perimeter of the window is 40 ft, find the dimensions of the window so that the greatest possible amount of light is admitted.
(a) Find the width of the window
(b) Find the height of the window.

Answer 1

The window is a rectangle with a semicircle sitting on top of it. If the semicircle has radius r, then the width of the rectangle is 2r.

If the width of the rectangle is 2r, and the perimeter is 40, then the height is:

p = 2h + 2w (formula for rectangle perimeter)
40 = 2h + 2(2r) (width = 2r)
2h = 40 - 4r
h = 20 - 2r

Now, the area of the entire window is the area of the rectangle plus the area of the semicircle
...

a = (rectangle area) + (semicircle area)
a = (w * h) + (pi * r^2/2)
a = ((20 - 2r) * (2r)) + (pi * r^2/2)
a = 40r - 4r^2 + (pi * r^2/2)
a = 40r + ((pi - 8)/2)r^2

To maximize the area, we find its derivative with respect to r, and set that equal to zero:

da/dr = 0
d(40r + ((pi - 8)/2)r^2)/dr = 0
40 + (pi - 8) * r = 0
r = 40 / (8 - pi)
r =~ 8.233

Now, the width = 2r, so that's twice about 8.233, or about 16.466.

And the height is 20 - 2r, or about 3.534.

Answer 2

Find the expression for area in terms of the width.
so if w is the width and l is the length we have

Area = A = lw + (pi)((w)^2)/8
as 2(l+w) = 40
so l = 20 - w

so A = (20-w)w + (pi)((w)^2)/8

differentiate with respect to w and equate to 0 for maximum value of Area.

so w = 80/(pi)-8)
and l = 20((pi) - 12)/((pi) - 8)

Theres ur answer.

Answer 3

i have were given also lengthy surpassed by potential of this and the algebra is a sprint daunting and that i can't upward push up with any type of short cut back to this difficulty. that is quite interesting software. you've finished perfect with it.