$$\begin{array}{ll} \text{minimize} & e^{x-y}\\ \text{subject to} & e^x + e^y \leq 20\\ & x \geq 0\end{array}$$
My attempt:
To minimize the objective function, minimise $(x-y)$, i.e., maximize $y$ and minimize $x$. Given the 2nd constraint, the minimal value of $x$ is $0$. So, $x_* = 0$. Using this in the first constraint, we obtain $e^y \leq 19$. To maximize $y$, maximize $e^y$, and so the first constraint holds with equality. $y_* = \ln (19)$. Hence, the minimum is $e^{x_* - y_*} = \frac{1}{19}$. Please tell me if it's correct.