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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 \begin{equation} x_1+2x_2+p_3x_3\le w \end{equation} The corresponding Lagrangian (where $ \mu $ is the Lagrange multiplier) is \begin{equation} L=\sqrt{x_1x_2}+c x_3+\mu(w-x_1-2x_2-p_3x_3). \end{equation} Taking F.O.Cs, we have \begin{equation} \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. \end{equation}

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

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  • $\begingroup$ Hi! 1. What do you mean when you say this system of equation has no solution? $\endgroup$
    – Giskard
    Aug 27, 2023 at 17:51
  • $\begingroup$ 2. Have you considered adding non-negativity constraints? $\endgroup$
    – Giskard
    Aug 27, 2023 at 17:51
  • $\begingroup$ 3. Are you intent on using a Langrangian or would some other method suffice as well? $\endgroup$
    – Giskard
    Aug 27, 2023 at 17:52

1 Answer 1

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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*}

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  • $\begingroup$ Thanks! This solution is really helpful. I am just wondering if there is any standard reasoning for solving it this way? Say like when facing a quasi-linear utility or some other weird things? $\endgroup$
    – Paul Huang
    Aug 28, 2023 at 5:26
  • $\begingroup$ This is a standard reasoning which I use in my lectures for solving such problems. To learn the details, please visit: youtube.com/… $\endgroup$
    – Amit
    Aug 28, 2023 at 6:35

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