Non-linear optimization problem using Lagrange's method/K.K.T. conditions

Question

We are given the following problem:

$$\text{minimize } 2x_1^2 + x_2^2 + 3x_3^2 \text{ subject to } x_1+x_2+x_3=10, x_1\le5, \text{ and } x_1,x_2,x_3\ge0$$

I understand that I have to check all possible combinations for $\lambda_1$ and $\lambda_2$ and then choose the one that matches the conditions. I am pretty sure that the correct one is $\lambda_1\gt0, \lambda_2=0$, however I cannot prove it.

This is my progress so far:

$L(x_1, x_2, x_3,z_1;λ_1, λ_2)=2x_1^2+x_2^2+3x_3^2+λ_1(x_1+x_2+x_3-10)+λ_2(x_1+z_1-5)$

K.K.T. Conditions:

i. $4x_1+λ_1+λ_2=0, 2x_2+λ_1=0, 6x_3+λ_1=0$

ii. $x_1+x_2+x_3=10, x_1\le5$

iii. $λ_1\ge0, λ_2\ge0$

iv. $λ_2(x_1-5)=0$

Answer 1

The KKT conditions are \begin{align} 4x_1+\lambda_1+\lambda_2 &=0\\ 2x_2+\lambda_1 &= 0\\ 6x_3+\lambda_1 &= 0\\ x_1+x_2+x_3 &= 10\\ x_1 &\le 5 \\ x_i &\ge 0 \\ \lambda_2 &\ge 0 \\ \lambda_2(x_1-5) &= 0\\ \end{align} The complementary slackness constraint implies two cases: $\lambda_2=0$ or $x_1=5$. The first case yields solution $x=(30/11,60/11,20/11)$, with objective value $600/11$. The second case yields solution $x=(5,15/4,5/4)$, with objective value $275/4>600/11$.

Answer 2

To facilitate the graphic representation we will eliminate the equality constraint so the problem becomes bi-dimensional. Now calling

$$ \cases{ f(x,y) = -(2x^2+y^2+3(10-x-y)^2)\\ g_1(x,y,s) = -x+5-s^2\\ g_2(x,y,s) = x-s^2\\ g_3(x,y,s) = y-s^2\\ g_4(x,y,s) = 10-x-y-s^2 } $$

we have the lagrangian

$$ L(x,y,\lambda,s) = f(x,y) + \sum_{k=1}^4\lambda_k g_k(x,y,s_k) $$

Here $s_k$ are slack variables to transform the due inequalities into equations. Now the stationary points are the solutions for

$$ \nabla L = \left\{ \begin{array}{l} \lambda_1+\lambda_4+10 x+6 y=\lambda_2+60 \\ \lambda_3+60=\lambda_4+6 x+8 y \\ \lambda_1 s_1=0 \\ \lambda_2 s_2=0 \\ \lambda_3 s_3=0 \\ \lambda_4 s_4=0 \\ s_1^2+x=5 \\ s_2^2=x \\ s_3^2=y \\ s_4^2+x+y=10 \\ \end{array} \right. $$

giving

$$ \left( \begin{array}{ccccccccc} f&x&y&\lambda_1&\lambda_2&\lambda_3&\lambda_4&s_1^2&s_2^2&s_3^2&s_4^2\\ 54.5455 & \frac{30}{11} & \frac{60}{11} & 0 & 0 & 0 & 0 & \frac{25}{11} & \frac{30}{11} & \frac{60}{11} & \frac{400}{121} \\ 66.6667 & \frac{10}{3} & \frac{20}{3} & 0 & 0 & 0 & -\frac{40}{3} & \frac{5}{3} & \frac{10}{3} & \frac{20}{3} & 0 \\ 68.75 & 5 & \frac{15}{4} & -\frac{25}{2} & 0 & 0 & 0 & 0 & 5 & \frac{15}{4} & \frac{25}{16} \\ 75. & 0 & \frac{15}{2} & 0 & -15 & 0 & 0 & 5 & 0 & \frac{15}{2} & \frac{25}{4} \\ 75. & 5 & 5 & -10 & 0 & 0 & -10 & 0 & 5 & 5 & 0 \\ 100. & 0 & 10 & 0 & -20 & 0 & -20 & 5 & 0 & 10 & 0 \\ 125. & 5 & 0 & 10 & 0 & -30 & 0 & 0 & 5 & 0 & 25 \\ 300. & 0 & 0 & 0 & -60 & -60 & 0 & 5 & 0 & 0 & 100 \\ \end{array} \right) $$

The next step is the correct qualification for those stationary points. From the attached graphics, are not feasible minimizers all the points for which according to KKT

$$ \nabla f(x^*,y^*) = \sum_{k\in \sigma(x^*,y^*)}\lambda_k\nabla g_k(x^*, y^*),\ \ \ \lambda_k \ge 0 $$

Here $\sigma(x^*,y^*)$ represent those restriction indexes active at the stationary point $(x^*, y^*)$. This is because as $\sum_{k\in \sigma(x^*,y^*)}\lambda_k\nabla g_k(x^*, y^*)$ represent a cone inside the feasible region, if $\nabla f(x^*,y^*)$ is contained in it, then $(x^*, y^*)$ could slide along $\nabla f(x^*,y^*)$ to decrement the objective function value.

As can be observed, the only feasible minimizers are the points located at $\left(\frac{30}{11},\frac{60}{11}\right)$ and $(5,0)$. Here in black $\nabla f(x^*,y^*)$ and in red $\nabla g_k(x^*,y^*)$