A 2-person society in which there are 2 consumption goods $x_1,x_2$. Individual 1's utility function is $u_1(x_1^1,x_2^1)=6+0.4ln(x_1^1)+0.6ln(x_2^1)$, while individual 2's utility is $u_2(x_1^2,x_2^2)=6+ln(x_1^2)+ln(x_2^2)$. The social endowment consists of 15 units of the irst good and 20 units of the second good, to be split between the two. What division of the social endowment between the two individuals maximizes the social utility functional $2min({u_1,u_2})+max({u_1,u_2})?$
Using numerical method, I discretize the endowments, and evaluate the social welfare on each grid points of the 1500*2000 matrix, where each step is 0.01. I find the social optimal solution to be where $x_1^1=6.67,x_1^2=15-6.67,x_2^1=10.91,x_2^2=20-10.91$. It gives the point $(u_1=8.192855,u_2=12.32)$
But how to find it analytically, without using computational methods? I am a little confused: for individual 1, utility maximization requires marginal utility of eating two goods to be same: $x_2^1/x_1^1=3/2$, while for individual 2, $x_2^2/x_1^2=1$. At the solution I find, these condition are not hold...I am a little rusty on GE and social choice, any hints?
Thanks in advance.
I don't like superscripts. So let me call $x_1^1=a_1$, $x_2^1=b_1$, $x_1^2=a_2$, $x_2^2=b_2$. Then we have
\begin{align} u_1(a_1,b_1)&=6+0.4\ln a_1+0.6\ln b_1,\\ u_2(a_2,b_2)&=6+\ln a_2+\ln b_2. \end{align}
We wish to maximize the utility function
\begin{align} U&=2\min(u_1,u_2)+\max(u_1,u_2)\\ &=u_1+u_2+\min(u_1,u_2), \end{align}
subject to the constraints $\,a_1+a_2=15$, $b_1+b_2=20$, and $\,a_1,a_2,b_1,b_2>0$. The constraints are convex and the objective function to maximize $\,U\,$ is concave (minimum of two concave functions is concave). So the whole problem is convex and has a unique optimum. Analytically you consider two cases $\,u_1\geq u_2\,$ and $\,u_2\geq u_1\,$, do them separately and see which case gets the answer. Since your numerical answer says $u_1$ is smaller, I'll only do the $\,u_2\geq u_1\,$ case, so $\,\min(u_1,u_2)=u_1$. Thus it's $\,2u_1+u_2\,$ that we maximize. We have
\begin{align} U&=2u_1+u_2=18+0.8\ln a_1+\ln a_2+1.2\ln b_1+\ln b_2\\ &=18+0.8\ln a_1+\ln(15-a_1)+1.2\ln b_1+\ln(20-b_1). \end{align}
The require $\,\partial U/\partial a_1=\partial U/\partial b_1=0\,$ to find
$$a_1=\frac{20}{3},\quad b_1=\frac{120}{11}.$$
Then you should check if $\,u_2\geq u_1\,$ really agrees with the assumption made at the very beginning. If yes, we have the answer (which agrees with your numerical results). If no, it means the other case $\,u_1\geq u_2\,$ has the answer. This is the procedure to solve the problem analytically.