This problem comes from Schaum's Outlines "Calculus for Business, Economics, and the Social Sciences". Problem 9.14 (c) states: Maximize $f(x,y,z)=3x^2yz$ subject to x+y+z=32
So, I set up the Lagrangian function as normal: $F = 3x^2yz + L(x + y + z - 32)$ Solve the partial derivatives (an easy problem) and you get x0=16, y0=8 and z0=8, and f=49,152.
This appears to work in the sense that if you change x,y,z by a small amount, f gets smaller. But I spotted the x-squared term, I wondered what would happen if x=-50, y=40 and z=42, in which case f is MUCH bigger and the constraint is met.
What am I missing?
Lagrange multipliers finds local extrema. On a domain with boundary, global extrema might be on the boundary instead. On an unbounded domain, the objective function might not be bounded above (resp. below), even on the surface where the constraint holds, in which case the maximum (resp. minimum) could be at infinity.
In this case, assuming the problem context is about $(x,y,z)$ in the whole space, I think you are right, and you can make $f$ arbitrarily large by considering, say, $(x,y,z)=(-t,t+16,t+16)$ for large $t>0$. But I wouldn't be surprised if there was some implicit assumption missing somewhere, for instance that $x,y,z \geq 0$.