Find two positive real numbers x and y such that they add to 120 and x^(2)y is as large as possible

Answer 1

x+y = 120, so y = 120-x

z = x^2y = x^2(120-x)
dz/dx = 240x - 3x^2 = 3x(80-x)
absolute maximum of x^2y is at x=80

so, x=80, y=40

Note that this is very like the fencing problems, where the length is divided equally among the lengths and widths. Here, the powers of x and y are the key as to how to divide the 120: 40 for x, 40 for x, 40 for y.