Solving this problem with Lagrange multipliers method Hi everyone I want to solve this problem with Lagrange multiplier method 
This is the problem 
$$
   \operatorname{Min} F(x)=3a x_1+5a x_2
$$
subject to these Constraints :
$$
g_1 : (2.16/x_1)+(10/x_2)\leq 1
$$
$$
g_2 : x_1 \geq 11.25
$$
$$
g_3 : x_2 \geq  18.75
$$

  
*
  
*(Note 1: I only know a little about KKT conditions for solving    nonlinear problems that has one inequality and one equality
  constraints and i only seen that in an example , so i don't know much 
  about method)
(Note 2: This problem is based on weight optimization of a symmetric truss (engineering design problem)  that i formulated that and Objective function
  F(x) and Constraints came from    optimization problem
  formulation method that uses finite element method    and reason i'm telling you this because
  i think you want to know the origin of my    question)

 A: Because $a$ is a positive constant we can ignore it and minimise
$$
F(x)=3 x_1+5 x_2
$$
Rewrite the inequalities with slack variables:
$$
\frac{2.16}{x_1}+\frac{10}{x_2}\leq 1 \implies \frac{2.16}{x_1}+\frac{10}{x_2}+s_1^2= 1\\
g_1=\frac{2.16}{x_1}+\frac{10}{x_2}+s_1^2-1
$$
Now the second one:
$$
x_1 \geq 11.25\implies x_1 = 11.25+s_2^2\\
g_2= x_1-11.25-s_2^2
$$
Finally, the third one:
$$
x_2 \geq  18.75 \implies x_2 = 18.75+s_3^2\\
g_3 = x_2 - 18.75-s_3^2
$$
Form the Lagrangian:
$$
\begin{split}
L(x_1,x_2,\lambda_1,\lambda_2,\lambda_3,s_1,s_2,s_3)
 &= 3 x_1 + 5 x_2 \\
 &- \lambda_1
    \left(\frac{2.16}{x_1}+\frac{10}{x_2}+s_1^2-1\right) \\
 &- \lambda_2 \left(x_1-11.25-s_2^2\right) \\
 &- \lambda_3 \left(x_2 - 18.75-s_3^2\right)
\end{split}
$$
This implies the following partial derivatives:
$$
\begin{split}
\frac{\partial L}{\partial x_1} &= 3+\frac{2.16\lambda_1}{x_1^2}-\lambda_2\\
\frac{\partial L}{\partial x_2} &= 5+{{10\lambda_1}\over{x_2^2}}-\lambda_3\\
\frac{\partial L}{\partial s_1} &= -2s_1\lambda_1\\
\frac{\partial L}{\partial s_2} &= 2s_2\lambda_2\\
\frac{\partial L}{\partial s_2} &= 2s_3\lambda_3
\end{split}
$$
