Multivariable Constrained Optimization QUESTION:
A vacuum cleaner company manufactures and sells three types of vacuum cleaners: the standard, executive, and deluxe models. The annual revenue (in dollars) as a function of the numbers x, y, and z (respectively) of standard, executive, and deluxe models sold is
$$R(x,y,z)=xyz^2-25000x-25000y-25000z$$
2
The manufacturing plant can produce up to (that is, less than or equal to) 200,000 total vacuum cleaners annually. 
So, the production function, P, would be $$P(x,y,z)=x+y+z$$ where $$x+y+z≤200000$$
Assuming that everything that is manufactured is sold, find how production should be distributed among the models so as to maximize the annual revenue?
My Work:
So far, I have solved for the interior critical points of f, by finding where $\nabla f=0$ $$\nabla f(x,y,z)=<yz^2-25000,xz^2-25000,2xyz-25000>$$
I then solved for x,y,z by creating a system of equations:$$yz^2-25000=0$$ $$xz^2-25000=0$$ $$ 2xyz-25000=0$$
Which has a solution at $$x=\dfrac{25000}{50000^{2/3}}\approx18.4202, y=\dfrac{25000}{50000^{2/3}}\approx 18.4202, z=\sqrt[3]{50000}\approx 36.8403$$
Which it clearly inside the constraint $x+y+z≤200000$.
At this point I got lost. These numbers seem far too low given the constraint at 200000, which makes me think I already did something wrong, and I don't know where to go from here.$$$$
 Would I then solve for the boundary points using Lagrange multiplier method by setting $\nabla f=\lambda*\nabla g$?
 A: As you point out there is only one (real) critical point of $R$, 
$$
(x, y, z) = (18.42, 18.42, 36.84)
$$
for which the function value attained is
$$
R(18.42, 18.42, 36.84) = -1.38 \cdot 10^{-6}.
$$
Clearly a negative revenue is not ideal and what we have found is in fact a local minimum of the function $R$. This leads us to believe that the maximum is attained at the boundary. We can investigate this by defining the Lagrangian
$$
\mathcal{L}(x, y, z, \lambda) = xyz^2 - 25000(x+y+z) + \lambda(200000 - x - y - z)
$$
for $\lambda \geq 0$. This allows us to define the first order necessary KKT conditions as 
$$
\begin{align}
\frac{\partial \mathcal{L}}{\partial x} = 0 \quad
\frac{\partial \mathcal{L}}{\partial y} = 0 \quad
\frac{\partial \mathcal{L}}{\partial z} = 0 \quad
\frac{\partial \mathcal{L}}{\partial \lambda} = 0
\end{align}
$$
which leads us to the very similar system of equations
$$
\begin{align}
yz^2 - 25000 - \lambda = 0 \\
xz^2 - 25000 - \lambda = 0 \\
2xyz - 25000 - \lambda = 0 \\
200000 - x - y - z = 0
\end{align}
$$
Notice that the latter is just the constraint. Solving this system is a bit of work so I let Wolfram Alpha do it for me. The system has six solutions, but one in which $\lambda \geq 0$. The KKT conditions specify that $\lambda \geq 0$ for the solution to be a maximum of the function. We fetch this single solution
$$
(x, y, z, \lambda) = (50000, 50000, 100000, 499999999975000)
$$
and evaluate our revenue function for this point
$$
R(50000, 50000, 100000) = 24999999995000000000.
$$
Now this is a revenue I can get behind! 
