$$\begin{array}{ll} \text{minimize} & e^{x-y}\\ \text{subject to} & e^x + e^y \leq 20\\ & x \geq 0\end{array}$$

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.

  • $\begingroup$ Your answer is correct ! $\endgroup$ – DeepSea Dec 4 '17 at 22:13
  • $\begingroup$ @LUCIFER If you are ok, you can set as solved. Thanks! $\endgroup$ – user Dec 6 '17 at 13:17
  • $\begingroup$ How do I do that? $\endgroup$ – LUCIFER Dec 6 '17 at 23:05
  • $\begingroup$ @LUCIFER Please, if you are ok, you can accept the answer and set it as solved. Thanks! Take a look here for instructions cdn.sstatic.net/img/faq/faq-accept-answer.png $\endgroup$ – user Jan 20 '18 at 22:36


As an alternative method to verify consider:


let $$a=e^{x} \quad b={e^y}$$

and study for $a\geq1$ and $b>0$


Here is a graphical interpretation of what is going on:

enter image description here


We have: $e^y \le 20 - e^x\implies e^{x-y}= \dfrac{e^x}{e^y}\ge \dfrac{e^x}{20-e^x}=\dfrac{20}{20-e^x}-1\ge \dfrac{20}{20-1}-1=\dfrac{1}{19}$, and this is the minimum which occurs when $x = 0, y = \ln 19$ .


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