Maximisation Problem (1 Viewer)

[skillstriker]

A window frame has the shape of a rectangle surmounted by a semicircle. The perimeter of the frame is constant. Show that, for maximum area, the height of the rectangle is equal to the radius of the semicircle.

Thanks!

 

[Aesytic]

referring to the image, perimeter is P = 2r + 2h + r*pi for some constant P
.'. 2h = P - 2r - r*pi
area is 2rh + (pi*r^2)/2 = r(P - 2r - r*pi) + (pi*r^2)/2
= Pr - 2r^2 - (pi*r^2)/2
to find max area, we differentiate this expression:
P - 4r - r*pi
.'. max area when P - 4r - r*pi = 0
P = 4r + r*pi
2r + 2h + r*pi = 4r + r*pi
2h = 2r
h = r
.'. max area when height of rectangle is the radius of the semi circle