Maxima: unable to solve Lagrangian I think this is may be more of a technical issue than a mathematical issue. I don't know...
I have the following function I want to maximize:
$$\arg\max_{x,y}\ U(x,y) = \frac{1-e^{-x}}{2} + \frac{1-e^{-y}}{2}$$
subject to $x+y \leq 50$ and $x,y\geq0$.
(My actual poblem is more complicated than this of course, but take this as a MWE please. :))
We can easily see this has a maximum. Using Maxima:
U(x,y) := (1-exp(-x))/2 + (1-exp(-y))/2;
wxplot2d([U(x,50-x)], [x,0,50]);


To solve this using Lagrangian optimzation along the $x+y=50$ frontier, I would do:
$$\begin{cases}
\dfrac{\partial U}{\partial x} = \lambda \\[4pt]
\dfrac{\partial U}{\partial y} = \lambda \\[4pt]
x+y = 50
\end{cases}
$$
We can easily solve this. $\frac{\partial U}{\partial x} = \frac{e^{-x}}{2}=\lambda \iff \frac{e^{-x}}{2} = \frac{e^{-50+x}}{2} \iff x=25$, as could be seen in the graph anyway.
But now I try to do this in Maxima:
U(x,y) := (1-exp(-x))/2 + (1-exp(-y))/2;
solve([diff(U(x,y),x)=lambda, diff(U(x,y),y)=lambda, x+y=50], [x,y,lambda]);

And it says there is no solution:
[]

What gives? I haven't done anything in optimization or Maxima for a long time... But I cannot for the life of me discover what the problem is...
EDIT: I would prefer to know where my Maxima usage is wrong (or my mathematics), but I will approve any answers that suggest me to use another symbolic solver.

Your solution is correct. It is also the same in Mathematica. One can still use "Reduce" in Mathematica as follows:
Reduce[{E^-x/2 - a == 0, E^-y/2 - a == 0, x + y == 50}, {x, y, a}]

with the following answer:
C[1] \[Element] Integers && (x == 25 + I \[Pi] + 2 I \[Pi] C[1] || x == 25 + 2 I \[Pi] C[1]) && y == 50 - x && a == E^-x/2

Since the problem is simple one can solve this problem without using any Lagrangian multiplier as well.
The function $U$ is increasing both in $x$, $y$ and in both. So this means for the maximization $x$ and $y$ should be as large as possible. So this means the constraint $x+y\leq 50$ should hold with equality, that is $x+y=50$. Then, one can simply insert $x:=-y+50$ in $U$, take the derivative w.r.t. $y$ and see that $y$ must be $25$. Accordingly $x=25$.
You may think about asking questions about programming in mathematica.stackexchange
