I know how to solve the two-good case with $u(x) = \min\{x1, x2\}$$u(x) = \min\{x_1, x_2\}$, but the addition of $x3$$x_3$ confuses me.
Problem
Derive the demand function $x(p,w)$ from $u(x) = \min\{x1, x2\} + x3$$u(x) = \min\{x_1, x_2\} + x_3$.
What I did so far
We assume that in optimum $x1 = x2$$x_1 = x_2$.
Set up the budget constraint $p1x1 + p2x2 + p3x3 = w$$p_1x_1 + p_2x_2 + p_3x_3 = w$.
Rewrite budget constraint as $(p1+p2)x1 + p3x3 = w$$(p_1+p_2)x_1 + p_3x_3 = w$ or $(p1+p2)x2 + p3x3 = w$$(p_1+p_2)x_2 + p_3x_3 = w$.
We can write $x1*$ or $x2* = \frac{w-p3x3}{p1+p2}$$x_1^*=x_2^* = \frac{w-p_3x_3}{p_1+p_2}$ and $x3* = \frac{w-(p1+p2)a}{p3}$$x_3^* = \frac{w-(p_1+p_2)x_1^*}{p_3}$.
Confusion
How to proceed? Can I still use a Lagrangian to solve this?
