Calculus Optimazation?

Question

This norman window is made up of a semicircle and a rectangle. The total perimeter of the window is 16 m. What is the maximum area?

Answer 1

Let a and b the lengths of the sides of the rectangle (with a the side with the semi-circle).
The perimeter is 2b+a(pi/2+1)=16, and the area ab+pi/8*a^2. Substituting b in the area (from the perimeter equation), the area is 8a-a^2*(1/2+pi/8).
Setting the derivative to 0, we have a=32/(4+pi). Then b comes from the perimeter constraint as b=8*(1-(2+pi)/(4+pi)).
Numerically this gives a=4.48, b=2.24 and an area of 17.92

Answer 2

Ok, start by finding an equation for the perimeter:
Here is the picture, click on it: http://img217.imageshack.us/img217/1149/optimization2gm0.jpg

the semicircle segment is half of the circumferance, or .5 x 2(pi)r, which comes out to just (pi)r

the perimeter of the rectangle would be the twice the radius time unknown distance x

((2r)x)+((pi)r)=P=16

The area would be .5 (pi)r^2+(2r)x=A

Use your "helper equation" from the perimeter, to solve for one of the variables, r or x
r(2x+pi)=16 is what the original equation for perimeter can be simplified to (by factoring out r)
r=16/(2x+pi)

plug this new value for r into the second equation for area, and then find the derivative of that, and then find when the derivative equals zero, and this x value will be at which the maximum area occurs

take this x value, and plug it into the original area equation to get your numerical value for maximum area

Answer 3

The length of the semicircle is pi*r, where r is radius.
The width of the rectangle is 2r.
The length of the window is [16 - (pir + 2r)]/2=8-pi*r/2 - r
So area =A = L X W = 2r[8-pi*r/2 - r] = 16r-pi r^2 -2r^2
dA/dr = 16 -2pi*r-4r = 16 -2r(pi-4)
Setting this = 0 gives r = 8/(pi-4)
The max area occurs when r = 8/(pi-4)
So Amax = 16(8/(pi-4))- pi(8/(pi-4)) - 2(8/pi-4)^2
=(128-pi)/(pi-4) -16/(pi-4)^2