Find the Nash equilibria for a joint project modeled as a strategic game

Question

I am currently working out of the Introduction to Game theory book by Martin J Osborne and I am looking to solve this given question:

Two people are engaged in a joint project. If each person $i$ puts in the effort $x_i$, a nonnegative number equal to at most $1$, which costs her $c(x_i)$, the outcome of the project is worth $f(x_1, x_2)$. The worth of the project is split equally between the two people, regardless of their effort levels. Formulate this situation as a strategic game. Find the Nash equilibria of the game when $$f(x_1, x_2) = 4x_1x_2$$ and $c(x_i) = x_i$ for $i = 1, 2$.

As such the general approach to this problem would be to take the pay off function, which in our case is $f$ and differentiate it with regards to $x_1$. Equate that to $0$ and solve for $x_1$ to make it a best response function for $x_1$. The same would be done for the $x_2$ variable and then we would have two equations, i.e. best response functions for both $x_1$ and $x_2$. We then solve these equations together and get the Nash Equilibria. The problem I'm having is when we differentiate $f(x_1,x_2)$ with respect to say $x_1$ we have $f' = 4x_2$ when we equate this to $0$ we have $0 = 4x_2$ which really doesn't help much in solving the problem... At this stage I'm really not sure how to go about solving this...

Here is a picture of the supposed solution:

Answer 1

The strategy of player 1 is to choose any $x \in [0,1]$. First, let's determine the best response function of player 1. By choosing $x_1$ player 1 pays $x_1$ and receives $f(x_1,x_2)$. Thus, the actual payoff (or utility) function of player 1 is equal to $$u_{1}(x_1,x_2)=\frac12f(x_1,x_2)-x_1=2x_1x_2-x_1=x_1(2x_2-1)$$ This is a linear function in $x_1$ whith slope parameter $2x_2-1$. Obviously $$2x_2-1=0 \implies x_2=\frac12$$ and therefore the best response of player 1 can be determined as follows

1. For $x_2<\frac12$. In that case $2x_2-1<0$ and therefore $u_1(x_1,x_2)$ is maximized for the least possible value of $x_1$, which is $x_1=0$. In symbols $$b_1(x_2)=0$$ for $x_2<\frac12$.
2. For $x_2=\frac12$. In that case $2x_2-1=0$ and therefore $u_1(x_1,x_2)=0$, so that $$b_1(x_2=\frac12)=x_1$$ for all $x_1\in[0,1]$.
3. For $x_2>\frac12$. In that case $2x_2-1>0$ and therefore $u_1(x_1,x_2)$ is maximized for the largest possible value of $x_1$, which is $x_1=1$. In symbols $$b_1(x_2)=1$$ for $x_2>\frac12$.

In sum $$b_1(x_2)=\begin{cases} 0, & \text{for } x_2<\frac12 \\ [0,1], & \text{for } x_2=\frac12 \\1, & \text{for } x_2>\frac12 \\ \end{cases}$$ and due to symmetry $$b_2(x_1)=\begin{cases} 0, & \text{for } x_1<\frac12 \\ [0,1], & \text{for } x_1=\frac12 \\1, & \text{for } x_1>\frac12 \\ \end{cases}$$ from which you can derive the given picture.