A rectangle is to be inscribed in a right triangle having sides of length 12in, 16in, and 20in.?

In the picture, replace 40 by 16 and 30 by 12 and 600 by 96

http://s1169.photobucket.com/user/chibuckt/media/maximum_area_rect_3_zpszzl5mff5.gif.html?filters[user]=127216769&filters[recent]=1&sort=1&o=0

Area of triangle = (1/2) (16)(12) = 96
96 = (1/2) (12) x + (1/2) (16) y
96 = area of triangle
96 = 6x + 8y
8y = 96-6x
y = (96-6x) / 8

Area = xy
Area = (1/8) x (96-6x)
Area = 12x-(6/8) x^2
Area = 12x - (3/4) x^2
dA/dx = 12 - (3/4) (2x) = 0
12 = (3/4) (2x)
12 = (6x / 4)
48 = 6x
x = 8
y = (96- 48) / 8 = 6

The dimensions of the rectangle are 6 (base ) and 8 (height)

d^2A/dx^2 = -6/4 < 0 ( so the area has been maximized)

No picture