Using Lagrange Multipliers to find minimum value of 3 variable equation Is possible to find minimum values of $f(x) = 7x +5y + 10z $  subject to the constraint $3x + y + 2z \geq 3$ and $-x + 2y + 4z \geq 5 $ ? I tried, and I get value of $\lambda $, but I think I made a mistake in the development. 
 A: The feasible domain of this problem is an infinite wedge, bounded by two halfplanes. The edge of this wedge is parallel to $(3,1,2)\times(-1,2,4)$, i.e. parallel to $v:=(0,-2,1)$. Since $\nabla f\cdot v=0$ the objective function $f$ is constant along this wedge, and this makes the problem somewhat degenerate. 
Inspection of the given data suggests to introduce the new variable $t:=y+2z$. In this way the dimension of the problem is reduced: Instead of $f(x,y,z)=7x+5(y+2z)$ we have to minimize the function $g(x,t):=7x+5t$ under the constraints
$$3x+t\geq3,\qquad-x+2t\geq5\ ,$$
or
$$3x+t=3+u,\qquad -x+2t=5+v\tag{1}$$
with "new coordinates"  $u\geq0$, $v\geq0$. Solving $(1)$ for $x$ and $t$ we obtain
$$x={1\over7}(1+2u-v)\qquad t={1\over7}(18+u+3v)\ ,$$
which leads to
$$\hat g(u,v)=g\bigl(x(u,v),t(u,v)\bigr)={1\over7}(97+29u+8v)\ .$$
As $u$ and $v$ have to be nonnegative the minimal possible value of $\hat g$, hence of the original $f$, is $${97\over7}\ .$$
A: The lagrangian method of multipliers should be done with equality restrictions so introducing two slack variables $\epsilon_i$ we have the lagrangian formulation
$$
L(x,y,z,\lambda_1,\lambda_2,\epsilon_1,\epsilon_2) = 7x+5y+10 z +\lambda_1(3x+y+2z-3-\epsilon_1^2)+\lambda_2(-x+2y+4z-5-\epsilon_2^2)
$$
The stationary points are the solutions for
$$
\nabla L = 0 =\left\{
\begin{array}{l}
 3 \lambda_1 -\lambda_2 +7 \\
 \lambda_1 +2 \lambda_2 +5 \\
 2 \lambda_1 +4 \lambda_2 +10 \\
 2 \epsilon_1 \lambda_1  \\
 2 \epsilon_2 \lambda_2  \\
 3 x+y+2 z-3 -\epsilon_1^2\\
 -x+2 y+4 z-5-\epsilon_2^2 \\
\end{array}
\right.
$$
Solving this system we obtain the solution
$$
x = \frac 17, y = \mu, z = \frac{1}{14}(18-7\mu), \lambda_1 = -\frac{19}{7}, \lambda_2 = -\frac 87, \epsilon_1 = \epsilon_2 = 0
$$
for $\mu\in \mathbb{R}$ giving $\frac{97}{7}$ as the minimum value.
From $\epsilon_1 = \epsilon_2 = 0$ we conclude that the two restrictions are actuating. Follows a graphic showing in blue a piece of the feasible region and in orange the plane $7x+5y+10z-15=0$

A: The primal:
$$f(x,y,z)=7x +5y + 10z \ \to \text{min} \\
\ \ \ \ 3x + y + 2z \geq 3\\
-x + 2y + 4z \geq 5$$
The dual:
$$g(a,b)=3a+5b \ \to \text{max} \\
3a-b \leq 7\\
a + 2b \leq 5\\
2a+4b\le 10$$
Maximum occurs at the corner point of the open feasible region:
$$g\left(\frac{19}{7},\frac87\right)=\frac{97}{7}.$$
Hence, the minimum of $f(x_0,y_0,z_0)$ is $\frac{97}{7}$. 
Can you find the critical point $(x_0,y_0,z_0)$ from the dual theorem?
