What combination of inputs results in the largest output? I'm trying to solve the following problem:

Production of a certain company depends on $3$ inputs $x,y,z$ in the following way:
$$f(x,y,z) = 50x^{\frac{2}{5}} y^{\frac{1}{5}} z^{\frac{1}{5}}$$
Budget of the company is $24 000$ dollars and products $x, y, z$ can be bought for $80$, $12$ or $10$ dollars per unit in that order. What combination of inputs results in biggest production?

It's a Calculus test preparation problem and I have no idea how I would start. Could you help me?
My idea is to use constrained extremas but I don't know how.
 A: 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 $$(x,y,z)=(150,500,600).$$ 
A: 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 $$x=150,y=500,z=600$$
A: 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 $(x,y,z)=(150,500,600)$.
A: You are trying to maximize $f(x, y, z) = 50x^{\frac{2}{5}} y^{\frac{1}{5}} z^{\frac{1}{5}}$ subject to $g(x,y,z) = 80x + 12y + 10z = 24000$. To make it simpler, where the maximum of $f$ is is the same as where the maximum of $h(x, y, z) = x^2yz$ is. You can use the method of Lagrange multipliers to solve this. 
We want to find $x, y, z$ such that $$\nabla h = \lambda\nabla g$$ where $\nabla h$ is the gradient of $h$. This then gives us the three equations 
$$2xyz = 80\lambda$$
$$x^2z = 12\lambda$$
$$x^2y = 10\lambda$$
From the constraint function, we also have that $$80x + 12y + 10z = 24000$$
Now that there are $4$ equations and $4$ variables, it is possible to solve for $x, y, z$, and $\lambda$. Although there are solutions where one of $x, y, z$ is $0$, these are extraneous because production would be $0$. The actual answer is then $$x = 150, y = 500, z = 600$$
