# Kunh-Tucker-Langrange - minimum points in $\text{int} (S)$

Let $$f(x,y,z) = x^3 - y^3 + z$$ and the constraint set $$S = \left\{ (x,y,z) \in \mathbb{R}^3 : x^2 + 2y^2 \leq 1, \, y = 2z \right\}$$ and assume that we want to find a $$\bar{x}$$ such that $$\min_\limits{x \in S} f(x) = f(\bar{x})$$.

Kuhn-Tucker-Lagrange:

Let $$f_0 = x^3 - y^3 + z, f_1 = x^2 + 2y^2 - 1, f_2 = y - 2z$$.

$$\begin{cases} \nabla f_0 + \lambda_1 \nabla f_1 + \lambda_2 \nabla f_2 = 0 \\ \lambda_1f_1 = 0 \\ f_2 = 0 \\ \lambda_1 \geq 0, \, \lambda_2 \in \mathbb{R} \end{cases}$$

• Case $$\lambda_1 = 0$$ :

we yield the points : $$x_1 = \left( 0, \frac{1}{\sqrt{6}}, \frac{1}{2 \sqrt{6}}\right)$$ and $$x_2 = \left( 0, -\frac{1}{\sqrt{6}}, -\frac{1}{2 \sqrt{6}}\right)$$

which are both in $$\text{int}(S)$$.

Question: Does the fact that we found these candidates in $$\text{int}(S)$$ ensures that every other point we find on the boundary of $$S$$ (Case $$\lambda_1 > 0$$ ) cannot be a minimum point (and thus $$x_2$$ is the minimum) and why?

• Since $$\frac{\partial f(x,y,z)}{\partial z}=1\neq 0$$ so there are no extrempoints. – Dr. Sonnhard Graubner Jun 21 '19 at 12:01
• @Dr.SonnhardGraubner Can you elaborate on that? – Paris Jun 21 '19 at 13:21

Making the substitution $$y\to 2z$$ the problem assumes the reduced form

$$\min f=x^3-8z^3+z\ \ \ \text{s. t.}\ \ \ g = x^2+8z^2-1 \le 0$$

and the lagrangian with the contribution of a slack vatiable $$s$$ is

$$L(x,z,\lambda,s) = f+\lambda(g+s^2)$$

The stationary conditions are

$$\nabla L = 0 = \left\{ \begin{array}{l} 3 x^2+2 \lambda x \\ -24 z^2+16 \lambda z+1 \\ s^2+x^2+8 z^2-1 \\ 2 \lambda s \\ \end{array} \right.$$

with solutions

$$\left( \begin{array}{ccccc} f & x & z & \lambda & s^2 \\ 0. & 0. & -0.353553 & -0.353553 & 0. \\ 0. & 0. & 0.353553 & 0.353553 & 0. \\ -0.136083 & 0. & -0.204124 & 0. & 0.816497 \\ 0.136083 & 0. & 0.204124 & 0. & 0.816497 \\ -1.02035 & -0.993481 & -0.0403049 & 1.49022 & 0. \\ 1.02035 & 0.993481 & 0.0403049 & -1.49022 & 0. \\ -0.00604972 & 0.22368 & -0.344595 & -0.335521 & 0. \\ 0.00604972 & -0.22368 & 0.344595 & 0.335521 & 0. \\ \end{array} \right)$$

Attached a graphic showing the feasible region with the objective function level curves, with the stationary points in red dots. The red vectors represent at the stationary points, the restriction gradient, as well as the black vectors, the objective function gradient. With those elements we can qualify the stationary points, choosing the solution set. 