What combination of inputs results in the largest output?

By AM-GM $$24000=80x+12y+10z=2(20x+20x+6y+5z)\geq8\sqrt[4]{20^2\cdot6\cdot5x^2yz}.$$ The equality occurs for $20x=6y=5z$ or $(x,y,z)=(150,500,600)$.

I.e., $x^2yz$ (and so $50(x^2yz)^{\frac{1}{5}}$) attains a maximal value for $$(x,y,z)=(150,500,600).$$

This is a standard Lagrange multiplier question. We want to maximize $$50x^{2/5}y^{1/5}z^{1/5}$$ subject to $80x+12y+10z=24000$.

So we try to maximise $$50x^{2/5}y^{1/5}z^{1/5}-\lambda(80x+12y+10z)$$ We know the maximum must be on the boundary or at a stationary point. But $xyz=0$ on the boundary, so it will be at a stationary point. Setting the three partial derivatives to 0 gives us three equations relating $x,y,z$ from which we easily deduce $y=10x/3,z=4x$. Substituting into the constraint then gives $$x=150,y=500,z=600$$

Let $g(x,y,z)=80x+12y+10z$ and use Lagrange multipliers: \begin{align} 20 x^{-3/5} y^{1/5} z^{1/5} &= 80 \lambda\\ 10 x^{2/5} y^{-4/5} z^{1/5} &= 12 \lambda\\ 10 x^{2/5} y^{1/5} z^{-4/5} &= 10 \lambda\\ 80x+12y+10z &= 24000 \end{align} The resulting solution is $(x,y,z)=(150,500,600)$.