Question

X+y+z= 158 , what’s the maximum value of xyz and when does that happen?

Answer 1

If you allow negative numbers there's no maximum.

If you restrict to values ≥ 0 then you can show the maximum is when x = y = z = 158/3, xyz = (158/3)^(3).

How to show that depends on the class. In algebra you might use arithmetic mean-geometric mean inequality. In calculus you might use Lagrange multipliers.

Answer 2

The clever way to show this is: by monotony, maximizing xyz is equivalent to maximizing log(xyz) = log(x) + log(y) + log(z). log() has decreasing slope so redistributing some of the x+y+z budget from a higher variable to a lower one always raises log(xyz) higher, so they must be equal.

Answer 3

This sounds like a job for ... Lagrange Multipliers!

Answer 4

What restrictions are on x, y and z - do they need to be positive integers, for example.