Maximize $xy^4$ subject to $xe^y \leq 3e^2,~ 0 \geq y,~ x \geq 0,~ y \geq 0$ I do not know how I am meant to complete this problem. It is intended to be completed without reference to the Kuhn-Tucker theorem or any other Lagrangian theorem or otherwise, I expect the solution to be simple and obvious, but I have nothing
All I can think of is just using the largest values I am allowed which is making $xe^y=3e^2$ where $(x,y)=(3,2)$, however, wolfram alpha gives an oddly specific answer $(x,y)=(0.406072,3.99984)$
 A: First, I assume that the $0 \geq y$ constraint in the title is a typo, since you also have $y \geq 0$, so I'm going to ignore it.
The crux here is that the question is not convex, but with a few transformations it can be made convex.
Specifically, the $x e^y \leq 3 e^2$ really wants to be $\log x + y \leq 2 + \log 3$, which suggests making the substitution $z = \log x$ and transforming the problem to
\begin{align*}
\mathrm{maximise}\quad & z + 4\log y \\
\mathrm{subject\;to}\quad & z + y \leq 2+\log 3,\\
&y \geq 0.
\end{align*}
(It's easy to see that the $xy^4$ will not be maximised at $x=0$, so we don't need to worry about that case).
But now, $z + 4\log y$ is increasing in both $z$ and $y$, so the maximum must occur on the boundary.
Since the objective is $-\infty$ on the $y = 0$ boundary, it must happen on $z + y = 2 + \log 3$, where the objective is
$$
z + 4\log y
= 4\log y - y + 2 + \log 3.
$$
This is smooth, and we can check using differentiation that it is maximised at $y = 4$, so that $z = \log 3 - 2$ and so $x = 3e^{-2} \approx 0.406$.
This basically agrees with Wolfram Alpha's computation, which used numerical methods and so wasn't exact.
