A window consists of an open rectangle topped by a semicircle and has a perimeter of 288 inches.?

Question

Find the radius of the semicircle that will maximize the area of the window. Please show work. Thanks. (:

Answer 1

Is it rectangle or a square..................???


window is surrounded by three sides and fourth side is a semicircle with radius 'r'..........
thus diameter of the semicircle=2r is the length of window

perimeter of semicircle=2*pi*r/2=pi*r

perimeter of window=2*breadth+length+pi*r
=2*b+r+pi*r
=2b+r(1+pi)

=>2b+r(1+pi)=288
=>r(1+pi)=288-2b
=>r=(288-2b)/(1+pi)..........answer

Now area of window=l*b+(pi*r^2)/2

Answer 2

If you let r to be the radius of the semicircle, and x be the length of the rectangle, then the Perimeter of the window is P where
P = pi*r + 2r + 2x
If A is the area of the window, then
A = pi * r^2 / 2 + rx
we know that P = 288, hence
288 = pi*r + 2r + 2x
=> x = (288 - pi*r - 2r)/2
substituting to x in r to the area
=> A(r) = pi* r^2 / 2 + r*((288 - pi*r - 2r)/2)
solving for the derivative,
=> A'(r) = pi*r + r(-pi/2 - 1) + (288 - pi*r - 2r)/2
equating to zero,
=> A'(r) = 0
=> r = 96

Answer 3

Radius = x
Diameter of semicircle = 2x
One side of rectangle = 2x
Each of other 2 sides = y

Perimeter = 2y + 2x + (pi)2x/2 = 288
2y + (2 + pi)x = 288
y = [288 - (2+pi)x]/2

Area, A = 1/2 pi x^2 + xy = 1/2 pi x^2 + x[288 - (2+pi)x]/2

To maximize
dA/dx = 0
1/2 pi 2x + [288 - (2+pi)x]/2 + x[0 -(2 + pi)]/2 = 0
multiply by 2
pi 2x + 288 - 2x - pi x -2x - pi x = 0
Solving, x = 72
Radius = 72"

Answer 4

Let

r = radius of semicircle
h = height of rectangle
p = perimeter
A = area

Given p = 288 in

Find r such that A is maximized.

We have

L = length = 2r

p = πr + 2h + L = πr + 2h + 2r = 288
V = πr²/2 + 2rh

Solving for h

p = πr + 2h + 2r = 288
2h = 288 - πr - 2r
h = 144 - πr/2 - r

Substituting for h

V = πr²/2 + 2rh = πr²/2 + 2r(144 - πr/2 - r)
V = πr²/2 + 288r - πr² - 2r²
V = 288r - (π/2 +2)r²

Take the derivative to find critical values.

dV/dr = 288 - (π +4)r = 0
288 = (π +4)r
r = 288/(π +4)

Take the second derivative to determine nature of critical value.

d²V/dr² = -π - 4 < 0 implies a relative maximum

r = 288/(π +4) ≈ 40.327139 in³