There is a question that I am trying to solve but I am not sure about my approach and is hoping I could get some help. Here is the question:
There are $n$ companies sharing a water reservoir, let's say the water pipe has a maximum transfer capacity 1 unit. Each company $i$ can receive $x_{i}$ units of water where $x_{i} \in [0, 1]$. Each company wants to take as much water as possible, but the water quality gets worse with the total water taken from the pipe. If the total water taken by all companies, $\sum_{i=1}^{n} x_{i}$ exceeds the maximum transfer capacity of 1 then the utility of each company is 0. The utility for company i is given by
$$u_i = \begin{cases} e^{x_i}\prod_{j=1}^{n} e^{{-x_i}{x_j}} -1, & \text{if } \sum_{i=1}^{n} x_{i} < 1, \\ 0, & \text{if } \sum_{i=1}^{n} x_{i} \geq 1. \end{cases}$$
We want to find the nash equilibrium and the social welfare of the equilibrium.
My idea of the equilibrium is the case where all players are at capacity of 1, so there is no incentive of any players to decrease, hence that is a Nash equilibrium. I am not sure if this is the right approach.
You found one Nash equilibrium. There is an entire class of similar Nash equilibria: Any strategy profile with $\sum_ix_i-x_j\ge1$ for all $j$ is a Nash equilibrium, since no company can individually reduce the consumption below capacity.
However, there is also a more interesting equilibrium with total consumption below capacity. Setting the derivative of the exponent of $u_i$ with respect to $x_i$ to $0$ yields
$$ \frac{\partial}{\partial x_i}\left(x_i-\sum_jx_ix_j\right)=1-\sum_jx_j-x_i=0\;, $$
or $x_i=1-\sum_jx_j$, and then summing over $i$ yields $\sum_ix_i=n\left(1-\sum_jx_j\right)$, with solution $\sum_ix_i=\frac n{n+1}$ and thus $x_i=\frac1{n+1}$. The social welfare in this case is
$$ n\left(\sum_i\mathrm e^{\frac1{n+1}-\sum_j\frac1{(n+1)^2}}-1\right)=n\left(\mathrm e^{\frac1{(n+1)^2}}-1\right)\approx\frac n{(n+1)^2}\;. $$
The maximum social welfare is achieved if a single company consumes with $x_i=\frac12$, and is given by
$$ \mathrm e^\frac14-1\approx0.284\;. $$