# Minimize $\frac{1}{2}(x_1^2+x_2^2)$ subject to $x_1-2x_2+2\le 0$

a) The Lagrangian is: $$L(x_1,x_2,{\lambda})=\frac{1}{2}(x_1^2+x_2^2) + {\lambda}\cdot (x_1-2x_2+2)$$

b) The KKT conditions are:

1) $x_1+{\lambda} = 0$

2) $x_2 - 2*{\lambda} = 0$

3) ${\lambda}*(x_1-2x_2+2) = 0$

c) If ${\lambda}$=0 is not possible, then ${\lambda}$>0 so the only possible minimizer is (-2/5,4/5), no?

How to study condition 2?

Thanks paul-henri

• In condition 3), you are deriving with respect to $\lambda$ so condition 3) should read $x_1-2x_2+2 = 0$. This does not change the rest of the problem.
– MasB
Commented Apr 24, 2017 at 21:49

This is not an answer the OP wants. I merely provide an alternative approach.

By AM-GM or Cauchy-Schwarz, we have \begin{align} x_1^2+x_2^2&=x_1^2+4\,\left(-\frac{1}{2}x_2\right)^2\geq \frac{1}{1+4}\,\Biggl(x_1+4\,\left(-\frac{1}{2}x_2\right)\Biggr)^2 \\ &=\frac{1}{5}\left(x_1-2x_2\right)^2 \geq \frac{1}{5}(-2)^2=\frac{4}{5}\,, \end{align} where the last inequality is due to $x_1-2x_2\leq -2$. The inequality $x_1^2+x_2^2\geq \frac{4}{5}$ becomes an equality if and only if $x_1=-\frac{1}{2}x_2$ and $x_1-2x_2=-2$, which is equivalent to $\left(x_1,x_2\right)=\left(-\frac{2}{5},\frac{4}{5}\right)$.

As you correctly state $\lambda = 0$ yields a contradiction. So by 3) we have

4) $(x_1-2x_2+2) = 0$

From 1) we know $x_1 = - \lambda$ and from 2) $x_2 = 2\lambda$. Replace in 4) to get

5) $(-\lambda - 4\lambda + 2) = 0$

from which $\lambda = 2/5$. Substituting back, $x_1 = - 2/5$ and $x_2 = 4/5$.

This is the only point that satisfies the necessary (first-order) condition. However, since the problem satisfies the assumption for convex programming (objective function is convex, constraint is linear), the first-order condition is sufficient and you know that you have found the global minimizer.

• And to study conditions order 2 ? Commented Apr 24, 2017 at 22:20