Constrained Optimization - Lagrange Multipliers (Example) Let $f(x, y, z) = xyz$
$h1(x, y, z) = x + y + z − 4$5
and $h2(x, y, z) = 2x − y$.
Goal:
Minimize $f(x, y, z)$ subject to $h1(x, y, z) = 0$ and $h2(x, y, z) = 0$. 
First part:  Show that every feasible point is regular.
Clear since $h1$ and $h2$ will always be linearly independent given that $z$ is zero.
Now:


*

*Use the first order necessary conditions to find all candidates for
local minimum points. 

*Compute the tangent spaces to all the candidates. 

*Use second order necessary and sufficient conditions to decide which of the points are indeed local minimum points.
I obtained the following system with $\lambda_1$ = lambda one and $\lambda_2$ = lambda two:
$yz - \lambda_1 - 2\lambda_2 = 0 $
$xz - \lambda_1 - \lambda_2 = 0$
$xy - \lambda_1 = 0$ 
$x+y+z-45 =0$
$2x-y =0$
Main questions:
Is this system correct? Will answering 2 involve simply determining $x,y,z$ in terms of $L1/L2$ and solving?  Is there a general procedure for answering parts 2 and 3?  If nothing else, how might one compute tangent spaces for candidates? 
Thank you for taking the time to read this rather long question.
 A: The Lagrangian of your problem is $L(x,y,z,\lambda_1, \lambda_2) = xyz - \lambda_1 (x+y+z-45) - \lambda_2 (2x-y)$ so your system of first-order conditions is almost correct: the second equation should be $xz - \lambda_1 + \lambda_2 = 0$.
I did not understand your justification that all feasible points are regular. Here you need to compare the gradients $\nabla h_1(x,y,z)$ and $\nabla h_2(x,y,z)$. In addition, they don't need to be linearly independent at all feasible points but only at solutions of the first-order conditions.
For each candidate $(x^*,y^*,z^*)$ the tangent space is defined as
$$
\mathcal{K}(x^*,y^*,z^*) := \left\{
d \mid \nabla h_i(x^*,y^*,z^*)^T d = 0, \ i = 1, 2
\right\}.
$$
It must be recomputed for each candidate. Finally, second-order necessary conditions require that the Hessian $\nabla^2 L(x^*,y^*,z^*,\lambda_1^*,\lambda_2^*)$ be positive semi-definite on $\mathcal{K}(x^*,y^*,z^*)$. This means that for all $d$ in $\mathcal{K}(x^*,y^*,z^*)$,
$$
d^T \nabla^2 L(x^*,y^*,z^*,\lambda_1^*,\lambda_2^*) d \geq 0.
$$
The second-order sufficient conditions require $> 0$ instead of $\geq 0$. Again, these conditions must be checked for each candidate. Note that the second derivatives of $L$ are only computed with respect to $x$, $y$ and $z$---the variables of the problem.
It's also possible that in some cases, the gradients $\nabla h_1$ and $\nabla h_2$ are not linearly independent but yet there exist solutions to the first-order conditions. In this case, there will be multiple possible choices of $\lambda^*$ and the second-order conditions take a slightly different form.
