An arched window formed from the union of a semicircle and a rectangle has total perimeter #540"cm"#. What is the maximum possible area of the window?

1 Answer
Jun 27, 2015

Answer:

Maximum area is achieved when the bounding rectangle is a square of side #1080/(pi+4)cm#

Explanation:

Let the length of the base of the rectangle be #t# and its height #h#.

Then the length of the perimeter of the semicircle is #pi/2t#

The total perimeter of the window is #pi/2t + t + 2h = 540cm#

So #2h = 540cm - (pi/2 + 1)t#

So #h = 270cm - (pi/4 + 1/2)t#

The area of the rectangle is #t * h = t * (270cm - (pi/4 + 1/2)t)#

The area of the semicircle is #1/2 pi (t/2)^2 = pi/8t^2#

So the total area of the window is

#t * (270cm - (pi/4 + 1/2)t) + pi/8t^2#

#=(pi/8 - (pi/4+1/2))t^2 + t*270cm#

#=t(270cm - (pi/8+1/2)t)#

This has maximum value when #t = 135/(pi/8+1/2)cm = 1080/(pi+4)cm#

#h = 270cm - (pi/4 + 1/2)t = 270cm - (1080(pi/4+1/2))/(pi+4)cm#

#=270cm-270(pi+2)/(pi+4)cm#

#=270((pi+4)-(pi+2))/(pi+4)cm#

#=270(2/(pi+4))cm#

#=540/(pi+4)cm = t/2#

The total height of the window is #h + t/2 = t#

Basically the maximum area is achieved when the bounding rectangle is a square of side #1080/(pi+4)cm#