# Question #2041c

##### 1 Answer

The maximum area is

radius of the semicircle is

#16/(4 + pi) (~~ 2.24)#

rectangular window is#16/(4+pi) xx 32/(4 + pi) (~~ 2.24 xx4.48)#

#### Explanation:

Let us set up the following variables:

# {(r, "Radius of the semicircle","(feet)"), (h, "Height of the rectangular window","(feet)"), (A, "Total area enclosed by the window", "(sq feet)") :} #

Our aim is to find

The total perimeter is that of

# 16 = (h + 2r + h) + (1/2)(2pir) #

# \ \ \ = 2r + 2h + pi r #

# :. 2h = 16 - 2r - pi r #

# :. \ \ h = 1/2(16 - 2r - pi r) #

And the total Area is that of a rectangle and a semicircle:

# A = (h)(2r) + (1/2)(pir^2) #

# \ \ \ = 2hr + 1/2 pi r^2 #

# \ \ \ = 2(1/2(16 - 2r - pi r))r + 1/2 pi r^2 #

# \ \ \ = 16r - 2r^2 - pi r^2 + 1/2 pi r^2 #

# \ \ \ = 16r - 2r^2 - 1/2 pi r^2 #

We now have the Area,

# (dA)/(dr) = 16 - 4r - pi r #

At a critical point we have

# 16 - 4r - pi r = 0 #

# :. \ \ \ \ 4r + pi r = 16 #

# :. \ \ \ r(4 + pi) = 16 #

# :. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \r = 16/(4 + pi) (~~ 2.24)#

With this value of

# A = 16(16/(4 + pi)) - 2(16/(4 + pi))^2 - 1/2 pi (16/(4 + pi))^2 #

# \ \ \ = 128/(4 + pi) (~~ 17.92)#

And:

# h = 1/2(16 - 2(16/(4 + pi)) - pi (16/(4 + pi))) #

# \ \ = 16/(4+pi) (~~ 2.24)#

We can visually verify that this corresponds to a maximum by looking at the graph of

graph{16x - 2x^2 - 1/2 pi x^2 [-5, 10, -5, 22]}

And also check that the perimeter is correct:

# P = 2r + 2h + pi r #

# \ \ = 2(16/(4 + pi)) + 2(16/(4 + pi)) + pi(16/(4 + pi)) #

# \ \ = 16 #