Check Kuhn-Tucker conditions

Question

How to check if $(0,1)$ point is the solution of this optimization problem using Kuhn-Tucker Theorem.

Find the min of $e^{x_1-x_2}-x_1-x_2$ where $x_1+x_2\le1,\ x_1\ge 0,\ x_2\ge0$

I am thinking about inverting last two conditions to $\le$ and using Lagrange multiplier.

Answer 1

\begin{align} \mathcal{L}&=e^{x-y}-x-y-\lambda_1(1-x-y)-\lambda_2(x)-\lambda _3(y)\\ \mathcal{L'}&=\binom{e^{x-y}-1}{-e^{x-y}-1} -\lambda_1\binom{-1}{-1}-\lambda_2\binom{1}{0}-\lambda _3\binom{0}{1}\\ \end{align} So you want to satisfy: \begin{align} 0=\mathcal{L'}&=\binom{e^{x-y}-1}{-e^{x-y}-1} -\lambda_1\binom{-1}{-1}-\lambda_2\binom{1}{0}-\lambda _3\binom{0}{1}\\ x+y&\leq1\\ x&\geq0\\ y&\geq0\\ \lambda_i&\geq0\\ \lambda_1(x+y-1)&=0\\ \lambda_2(x)&=0\\ \lambda_3(y)&=0 \end{align} At $(0,1)$,

Feasibility Checks Out

\begin{align} \lambda_3&=0\qquad\text{ From Complimentarity Equation}\\ \lambda_1&=1.36\qquad\text{ From Optimality Equation}\\ \lambda_2&=0.73\qquad\text{ From Optimality Equation}\\ \end{align}

Everything Checks Out.

It is a KKT point.

Similarly, you can check the Second Order Necessary Conditions.

Note : It is Optimal.