KKT conditions with two inequality constrains

Question

I need to minimize $f(x,y,z)=x^{2}+2y^{2}+3z^{2}$ subject to \begin{align*} &x-y-2z\leq 12\\ &x+2y-3z\leq 8. \end{align*} So I wrote the lagrangian of $f$. \begin{align*} \mathcal{L}(x,y,z,\lambda,\mu)=x^{2}+2y^{2}+3z^{2}-\lambda(x-y-2z)-\mu(x+2y-3z) \end{align*}

I am not sure if the KKT conditions I wrote are right.

$\frac{\partial \mathcal{L}}{\partial x}=2x-\lambda-\mu=0\\ \frac{\partial \mathcal{L}}{\partial y}=4y+\lambda-2\mu=0\\ \frac{\partial \mathcal{L}}{\partial z}=6z+2\lambda+3\mu=0\\ \lambda(12-x+y+2z)=0\\ \mu(8-x-2y-3z)=0 \\ \lambda,\mu \geq 0\\$

Are the above kkt conditions correct ? Am I missing any conditions? If these are correct I plan to solve the six equations to find the minimum. Could someone tell me if I am doing this right. Thank you.

Answer 1

It is correct. Just a quick reminder: Note that, in general, for inequality constraint $g(x) \leq 0$, we would consider $g(x)\mu = 0$. Hence, we should have $$\lambda(x-y+2z-12)=0\\ \mu(x+2y+3z-8)=0 .\\$$ But that is the same as what you have written down.

Answer 2

You could note that $(0,0,0)$ is a global minimum of $f$ and is strictly feasible.

Hence the multipliers will be zero at the minimum.