# Utility Maximization of a quasi-linear utility function

I am dealing with a quasi-linear utility function. For example $$U=(x_1x_2)^{0.5}+cx_3$$ with constrain $$w\ge x_1+2x_2+px_3$$.By taking c, w and p as constant, I function that by using Lagrange multiplier method the F.O.C equations does not have a solution. I just don't know what's the problem. Thanks!

My attempt:

The budget constraint under this case is $$$$x_1+2x_2+p_3x_3\le w$$$$ The corresponding Lagrangian (where $$\mu$$ is the Lagrange multiplier) is $$$$L=\sqrt{x_1x_2}+c x_3+\mu(w-x_1-2x_2-p_3x_3).$$$$ Taking F.O.Cs, we have \left\{ \begin{aligned} \frac{1}{2}x_2(x_1x_2)^{-\frac{1}{2}}-\mu=0 \\ \frac{1}{2}x_1(x_1x_2)^{-\frac{1}{2}}-2\mu=0 \\ c-p_3\mu=0 \\ w-x_1-2x_2-p_3x_3=0 \end{aligned} \right.

However, the above set of equations has no solution...

• Hi! 1. What do you mean when you say this system of equation has no solution? Aug 27 at 17:51
• 2. Have you considered adding non-negativity constraints? Aug 27 at 17:51
• 3. Are you intent on using a Langrangian or would some other method suffice as well? Aug 27 at 17:52

This is the problem we want to solve: $$\begin{eqnarray*} \max_{x_1,x_2,x_3} & x_1^{0.5}x_2^{0.5}+cx_3 \\ \text{s.t.} &x_1+2x_2+px_3\leq w \\ \text{and }& x_1\geq 0, x_2\geq 0, x_3\geq0\end{eqnarray*}$$ where $$c>0, p>0, w>0$$ are given. It can be re-written as: $$\begin{eqnarray*} \max_{0\leq x_3\leq w/p} \ \ \max_{x_1\geq 0,x_2\geq 0} & x_1^{0.5}x_2^{0.5}+cx_3 \\ \text{s.t.} & \ x_1+2x_2\leq w-px_3 \end{eqnarray*}$$ We can solve the problem in two steps. When we solve this: $$\begin{eqnarray*} \max_{x_1\geq 0,x_2\geq 0} & x_1^{0.5}x_2^{0.5}+cx_3 \\ \text{s.t.} & \ x_1+2x_2\leq w-px_3 \end{eqnarray*}$$ we get:
$$x_1=\dfrac{w-px_3}{2}$$ and $$x_2=\dfrac{w-px_3}{4}$$. Therefore, we can write the second step of the problem as: $$\begin{eqnarray*} \max_{0\leq x_3\leq w/p} & \dfrac{w-px_3+2\sqrt{2}cx_3}{2\sqrt{2}}\end{eqnarray*}$$
and the solution satisfy: $$\begin{eqnarray*} x_3\in \begin{cases} \left\{\dfrac{w}{p}\right\} & \text{if } 2\sqrt{2}c > p \\ \left[0, \dfrac{w}{p}\right] & \text{if } 2\sqrt{2}c = p \\ \left\{0\right\} & \text{if } 2\sqrt{2}c < p \end{cases}\end{eqnarray*}$$
Given $$x_3$$ chosen as above, the corresponding values of $$x_1, x_2$$ are $$\begin{eqnarray*} (x_1,x_2)= \begin{cases} (0,0) & \text{if } 2\sqrt{2}c > p \\ \left(\dfrac{w-px_3}{2},\dfrac{w-px_3}{4}\right) & \text{if } 2\sqrt{2}c = p \\ \left(\dfrac{w}{2},\dfrac{w}{4}\right) & \text{if } 2\sqrt{2}c < p \end{cases}\end{eqnarray*}$$