Calculate the $\lambda^{*}$ in the given lagrangian problem

Question

The associated Lagrangian of the optimization problem $$\underset{x, y \geq 0} \max [3x + 2y] \text{ subject to } 2x+3y \leq 1$$ is $\mathcal{L}(x,y,\lambda) = 3x + 2y + \lambda (1 - 2x - 3y)$ where $\lambda \geq 0$. How do I find the solution to this given by $(x^{*}, y^{*}, \lambda^{*})$?

I understand that $(x^{*}, y^{*}) = (0.5, 0)$. What I don't understand is how to calculate $\lambda^{*}$. In fact, I never cared about the $\lambda$ until now.

Answer 1

With $f = 3x + 2y $ the lagrangian reads

$$ \mathcal{L}(x,y,\lambda,s) = f + \lambda_1 (1 - 2x - 3y-s_1^2) + \lambda_2(x-s_2^2)+\lambda_3(y-s_3^2) $$

The stationary points are the solutions for

$$ \mathcal{L}=0=\cases{3-2\lambda_1+\lambda_2\\ 2-3\lambda_1+\lambda_3\\ 1 - 2x - 3y-s_1^2\\ x-s_2^2\\ y-s_3^2\\ \lambda_is_i} $$

which are

$$ \left[ \begin{array}{ccccccccc} f&x&y&\lambda_1&\lambda_2&\lambda_3&s_1^2&s_2^2&s_3^2\\ 0 & 0 & 0 & 0 & -3 & -2 & 1 & 0 & 0 \\ \frac{2}{3} & 0 & \frac{1}{3} & \frac{2}{3} & -\frac{5}{3} & 0 & 0 & 0 & \frac{1}{3} \\ \frac{3}{2} & \frac{1}{2} & 0 & \frac{3}{2} & 0 & \frac{5}{2} & 0 & \frac{1}{2} & 0 \\ \end{array} \right] $$

The point qualification is left to the reader...

Answer 2

$\lambda$ is like a tool that helps you finding the max./min. So if you can find $(x,y)$ without finding $\lambda$, I think $\lambda$ will become useless then. Actually there does exist some optimization problems that need the help of $\lambda$ but won't know what's the value of it.