# 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)

• What denotes $a$ here? – Dr. Sonnhard Graubner Jun 5 at 13:32
• it's a constant that has a value that changes from one problem to another and it's not defined .....if it helps you can give it like 2 or 3 but i need an answer with those if it's possible – Ali Eslami Jun 5 at 13:37
• Can we assume that $a$ is positive? – Dr. Sonnhard Graubner Jun 5 at 13:42
• yes definitely it's a positive constant because it's a design parameter – Ali Eslami Jun 5 at 13:54
• I get $$x_1=\frac{45}{4},x_2=\frac{75}{4}$$ – Dr. Sonnhard Graubner Jun 5 at 13:59

Because $$a$$ is a positive constant we can ignore it and minimise $$F(x)=3 x_1+5 x_2$$
$$\frac{2.16}{x_1}+\frac{10}{x_2}\leq 1 \implies \frac{2.16}{x_1}+\frac{10}{x_2}+s_1^2= 1\\ g_1=\frac{2.16}{x_1}+\frac{10}{x_2}+s_1^2-1$$ 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: $$\begin{split} \frac{\partial L}{\partial x_1} &= 3+\frac{2.16\lambda_1}{x_1^2}-\lambda_2\\ \frac{\partial L}{\partial x_2} &= 5+{{10\lambda_1}\over{x_2^2}}-\lambda_3\\ \frac{\partial L}{\partial s_1} &= -2s_1\lambda_1\\ \frac{\partial L}{\partial s_2} &= 2s_2\lambda_2\\ \frac{\partial L}{\partial s_2} &= 2s_3\lambda_3 \end{split}$$