Optimization...?

Question

If a Norman window is made up of a semicircle and a rectangle (a rectangle and half a circle on top) and the total perimeter of the window is 16 m how do I find the maximum area of the window?

Answer 1

The width of the rectangle and the diameter of the circle will be the same. The half circumference will be 0.5*pi*W and the Width's contribution to the perimeter will be 2*W. Whatever is left (twice the heigth of the window) is 2*H = 16 - (2*W + 0.5*pi*W)

Now you can build an equation for the area as a function of W.

Answer 2

There are two formulas that surround this problem: perimeter and area. The area is what we want to maximize, while we're given some form of the perimeter (since we're given the perimeter is 32).

The first thing to do is assign variables. Since we have a semicircle and a rectangle,
Let L = length of the rectangle, W = width of the rectangle.
It follows that, if the semicircle is sitting on top of the rectangle, the diameter of the semicircle is W.

Now, we develop a formula for the perimeter. The perimeter would consist of (1) three sides of the rectangle, and (2) half of a circle (since it's a semicircle). That is

P = (three sides of rectangle) + (circumference of semicircle)

The three sides of the rectangle would be L + W + L.
The circumference of the semicircle would be half the circumference of its corresponding circle. Remember that the formula for the circumference of a circle is pi x diameter. Therefore, our diameter is W, and we have HALF of the circumference. Therefore,

1/2 of circle's circumference = 1/2 of (pi x diameter)
= 1/2 (pi)(W) = (pi/2) W

So our perimeter is then

P = L + W + L + (pi/2)W

Grouping like terms, we have

P = 2L + W + (pi/2)W

Grouping the W terms, we get

P = 2L + (1 + pi/2)W
P = 2L + ([2 + pi]/2)W

And we're given P = 32, so

32 = 2L + ([2 + pi]/2)W

The area of the window would be given by

A = (area of rectangle) + (area of semicircle)

Area of a circle is pi(r^2). Since r = D/2,
Area of a circle is pi(D/2)^2, or pi(D^2 / 4), or (pi/4)D^2
But we want half of it, so we take 1/2 of it.
1/2 (pi/4)D^2 = (pi/8) D^2.

And D = W, so we have (pi/8) W^2

A = LW + (pi/8) W^2

From the first equation, 32 = 2L + ([2 + pi]/2)W, we can derive L, and get:

L = 16 - ([2 + pi]/4)W

Plugging this into our area equation, we have

A = [16 - ([2 + pi]/4)W]W + (pi/8)W^2
A = 16W - ([2 + pi]/4)W^2 + (pi/8)W^2

A = 16W - (2(2 + pi)/8 + pi/8)W^2
A = 16W - ( (4 + 2pi + pi)/8 )W^2
A = 16W - ( (4 + 3pi)/8 )W^2

This will be our area function, A(W)
A(W) = 16W - ( (4 + 3pi)/8 )W^2

To maximize the area, take the derivative.

A'(W) = 16 - ( (4 + 3pi)/4 )W

{Note; when taking the derivative of the second term, the 2 went down and caused the 8 in the denominator to turn into the 4).

Make A'(W) = 0,

0 = 16 - ( (4 + 3pi)/4 )W

( (4 + 3pi)/4 )W = 16

Multiplying both sides by 4/(4 + 3pi), we get

W = 64/(4 + 3pi)

This is WHERE the maximum occurs. WHAT the maximum actually is, is determine by plugging this value into our newly formed area function.

A( 64 / (4 + 3pi) ) = 16( 64 / (4 + 3pi) ) -
( (4 + 3pi)/8 ) ( 64 / (4 + 3pi) )^2

= 1024 / (4 + 3pi) - ( (4 + 3pi)/8 ) ( 4096 / (4 + 3pi)^2 )

= 1024 / (4 + 3pi) - (1/8) (4096 / (4 + 3pi) )

= 1024 / (4 + 3pi) - 512 / (4 + 3pi)

= 512 / (4 + 3pi)

Answer 3

The whole area is S=S1+S2, where S1=2rh is rectangular part, S2=pi*r^2/2 is semicircle part, r is radius, 2r is width, h is height of rectangular part, h+r being total height; thus perimeter p= 2r +h +pi*r +h = 2r+2h+pi*r;
So dS/dr =dS1/dr +dS2/dr = (2rh’ +2h) +pi*r = 0 demand; while h=p/2 –(1+pi/2)*r, hence h’= -(1+pi/2); thus S’=0 = (-2r –pi*r) +(p- 2r –pi*r) +pi*r =p – (4+pi) *r, thence r = p /(4+pi) = 2.24 and h= 8-2.5708*2.24 =2.24;
S’’ = 4h’ +pi = -4 –2pi +pi <0 means max! S=17.923;