Pre cal #11?

Question

A Norman window has the shape of a rectangle surmounted by a semicircle of diameter equal to the width of the rectangle. If the
perimeter of the window is 20 ft, what dimensions will admit the most light (i.e. maximum
area).

Answer 1

The area you want to maximize is:

H*W + (1/2)pi*(W/2)^2

H*W is the area of the rectangular portion
(1/2)pi*(W/2)^2 is the area of the semicircle

We know that the perimeter is equal to 20 feet.

perimeter = 2H + W + pi*W/2 = 20

2H + W is the perimeter of the rectangular portion
The circumference of a circle is 2piR and R=W/2 and we want 1/2 of this since the top is a semicircle.
pi*W/2 is the perimeter of the semicircular portion

Solve for H: H = 10 - (W/2)(1 + pi/2)

A = H*W + (1/2)pi*(W/2)^2
A = (10 - (W/2)(1 + pi/2))*W + (1/2)pi*(W/2)^2
A = (10 - W/2)W - (pi/4)W^2 + (1/8)pi*W^2
A = 10W - (W^2)/2 - (pi/8)W^2

dA/dW = 0 = 10 - W - (pi*W)/4
W = 10/(1 + pi/4)
W = 5.6
H = 2.8

Answer 2

Let x be length and y be width.
Then perimeter = 20=2x+y+((2pi(y/2))/2) =2x+y+((pi*y)/2).
And we want to maximize xy+((pi*(y/2)^2)/2) =
xy+((pi*y^2)/8).
So, from first equation, 20-y-((pi*y)/2)=2x,
or 10-(y/2)-((pi*y)/4)=x.
Substitute that in first equation,
area = 10y-(y^2/2)- ((pi*y^2)/4)+((pi*y^2)/8).

SO, take derivative and set to 0,
10-y-((pi*y)/2)+((pi*y)/4)=0.
SO,10=y(1+((pi)/2)-((pi)/4))).
So, y=10/(1+((pi)/2)-((pi)/4)) = 5.6.
Then x=10-(5.6/2)-((pi*5.6)/4) = 2.8.

SO, length = 2.8 and width=5.6 approximately.
But check calculations just in case.