Find the area of the largest rectangle with one corner at the origin, the opposite corner in the first

Question

Find the area of the largest rectangle with one corner at the origin, the opposite corner in the first quadrant on the graph of the line f(x) = 12-2x and sides parallel to the axes. 

The maximum possible area is

Answer 1

Area = l*w = (x)(12 - 2x) or:

A(x) = 12x - 2x²

Maximize the area by taking the derivative and setting it equal to 0

A'(x) = 12 - 4x

0 = 12 - 4x

x = 3

The the base is 3 and the height is 12 - 2(3) = 6

So the maximum area = (3)(6) = 18