KKT condition for linear equality

Question

I'm a bit confused regarding the stationarity condition of the KKT conditions.

Consider: $$\max_{x_1, x_2, 2x_1 + x_2 = 3} x_1 + x_2$$

From the stationarity condition, we know that there has to exist $\lambda$ s.t:

$$1 - 2\lambda = 0 \land 1 - \lambda = 0$$

which cannot hold. I must be missing something obvious. Thanks.

Answer 1

Define:

$$ f(x) = x_1 + x_2 $$

$$ g(x) = 2 x_1 + x_2 - 3 $$

The constrained optimization problem is:

$$ \mbox{Maximize} \ f(x) $$ $$ \mbox{Subject to} \ \ g(x) = 0 $$

Define the Lagrangian Function $$ L(x) = f(x) - \lambda g(x) = x_1 + x_2 - \lambda ( 2 x_1 + x_2 - 3 ) $$

Necessary conditions for optimality are: $$ (1) \ \ {\partial L \over \partial x_1} = 0 $$ $$ (2) \ \ {\partial L \over \partial x_2} = 0 $$ $$ (3) \ \ {\partial L \over \partial \lambda} = 0 $$

Simplifying, we get $$ (1) \ \ 1 - 2 \lambda = 0 \ \ \mbox{or} \ \ \lambda = {1 \over 2} $$ $$ (2) \ \ 1 - \lambda = 0 \ \ \mbox{or} \ \ \lambda = 1 $$ $$ (3) \ \ 2 x_1 + x_2 - 3 = 0 $$

Since (1) and (2) contradict each other, the necessary condition for optimality is not satisfied.

This means that the constrained optimization problem does not have a maximum value.

This is also clear geometrically.

We are trying to maximize the function $$ f(x) = x_1 + x_2 $$ along the straight line $$ 2 x_1 + x_2 = 3 $$

We notice that the sum of the coordinates of the points lying on the straight line $$ 2 x_1 + x_2 = 3 $$ can increase in its value arbitrarily. $2 x_1 + x_2 = 3$" />