Check Kuhn-Tucker conditions 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.

\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.