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Question: An agent who consumes three commodities has a utility function given by:

$u(x_1,x_2,x_3)=x^{1/3}_1+\min\{ x_2,x_3\}$

Given an income $I$, and prices of $p_1,p_2,p_3$. Describe the consumer’s utility maximization problem. Can the weierstrass and the Kuhn-Tucker theorems be used to obtain and characterize a solution? Why or why not?

attempt: I assume $x_i$ represents the quantity and belongs to $\mathbb R_{+}$. You can form the constraints as follows: $$ x_i \geq0 \quad\forall i \in [3] \\ \sum_{i=1}^3p_ix_i \leq I $$ You can simplify the objective by noting that for the utility to be the maximum. Hence, the final problem becomes,

$$\ \max_{x_1, x_2, x_3}x_1^{1/3} +x_2 \quad s.t. \\ x_i \geq0 \quad\forall i \in [3] \\ \sum_{i=1}^3p_ix_i \leq I \\ x_2=x_3 $$ Let's eliminate $x_3$ as we know that $x_2=x_3$. The problem simplifies to $$\ \min_{x_1,\ x_2}\ -x_1^{1/3} -x_2 \quad s.t. \\ x_1 \geq0,\ x_2 \geq 0 \\ p_1x_1 + (p_2 + p_3)x_2 \leq I $$ $$\ \mathcal L(x_1, x_2)=-x_1^{1/3} -x_2 + \lambda_1(-x_1) + \lambda_2(-x_2) + \lambda(p_1x_1 + (p_2 + p_3)x_2 - I) $$

Comment: I am unsure how to take this further. I keep messing up the derivatives ( I assume) and when I try to solve for lambda I manage to fail to isolate the lambda variable let alone getting the variables x1,x2,x3. My professor encouraged me to try this complex problem as an “exercise for the reader.” How do I carry this further, or can someone show me a step by step solution from this point onwards?

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  • $\begingroup$ Hi: I must be missing something but why does $x_2 = x_3$. Is that a given constraint ? $\endgroup$
    – mark leeds
    Jun 13, 2020 at 22:10
  • $\begingroup$ Basically $x_2=x_3 $ because that’s the only way a max or min can hold. $\endgroup$
    – Tony456
    Jun 13, 2020 at 22:11
  • $\begingroup$ "by noting that for the utility to be the maximum." - something seems to be missing here. $\endgroup$
    – VARulle
    Jun 14, 2020 at 0:35
  • $\begingroup$ I'm not questioning you but just trying to understand. Are you saying that the only way to maximize the minimum of $x_2$ and $x_3$ is to set them equal ? If so, I get it. Thanks. $\endgroup$
    – mark leeds
    Jun 14, 2020 at 1:46
  • $\begingroup$ Hello, yes, that is what I am saying. We are given nothing else about the two variables and their relationship and as such the minimum of both of them have to the same number. $\endgroup$
    – Tony456
    Jun 14, 2020 at 3:00

3 Answers 3

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I assume $x_i$ represents the quantity and belongs to $\mathbb R_{+}$. You can form the constraints as follows: $$ x_i \geq0 \quad\forall i \in [3] \\ \sum_{i=1}^3p_ix_i \leq I $$ You can simplify the objective by noting that for the utility to be the maximum, $x_2 =x_3$. Try to reason why this is true. Hence, the final problem becomes,

$$\ \max_{x_1, x_2, x_3}x_1^{1/3} +x_2 \quad s.t. \\ x_i \geq0 \quad\forall i \in [3] \\ \sum_{i=1}^3p_ix_i \leq I \\ x_2=x_3 $$ Can you now set up the lagrange multiplier function? As far as KKT conditions are concerned, I believe they should help because you are trying to maximize a concave function(in $\mathbb R_+$) subject to convex constraints and the strict inequality constraint is linear.

Edit 1: Let's eliminate $x_3$ as we know that $x_2=x_3$. The problem simplifies to $$\ \min_{x_1,\ x_2}\ -x_1^{1/3} -x_2 \quad s.t. \\ x_1 \geq0,\ x_2 \geq 0 \\ p_1x_1 + (p_2 + p_3)x_2 \leq I $$ $$\ \mathcal L(x_1, x_2)=-x_1^{1/3} -x_2 + \lambda_1(-x_1) + \lambda_2(-x_2) + \lambda(p_1x_1 + (p_2 + p_3)x_2 - I) $$ Edit 2: I see solving the Lagrangian can be tough. Let's simplify, we know that all constraints cannot be active at the same time. It will yield an utility of zero.

Case-1: $x_1=0, x_2>0$ You can check algebraically, $x_2=\frac{I}{p_2+p_3}$. Hence, utility$(U)=\frac{I}{p_2+p_3}$

Case-2: $x_1>0, x_2=0$ You can check algebraically, $x_1=\frac{I}{p_1}$. Hence, utility$(U)=\left(\frac{I}{p_1}\right)^{1/3}$

Case-3 $x_1>0, x_2>0$ You can now use the Lagrangian, its much simpler than the original Lagrangian problem because $\lambda_1=\lambda_2=0$ $$\ \mathcal L(x_1, x_2)=-x_1^{1/3} -x_2 + \lambda(p_1x_1 + (p_2 + p_3)x_2 - I)\\ \frac{\partial\mathcal L}{\partial x_1} = 0 \implies x_1 = \left(\frac{1}{3\lambda p_1}\right)^{3/2}\\ \frac{\partial\mathcal L}{\partial x_2} = 0 \implies \lambda = \frac{1}{p_2+p_3}\\ \text{Use the constraint, }p_1x_1 + (p_2 + p_3)x_2 = I \text{ to find } x_2 $$ You can simply finish saying the solution is $\max$ of all three cases. As per the KKT conditions are concerned, one of the three solutions will satisfy KKT Conditions depending on $p_1, p_2, p_3 \text{ and } I$. Note that this is no different from the original problem and hence, KKT condition should be satisfied by one of the three depending on $p_1, p_2, p_3 \text{ and } I$. We could have obtained the same solution using the original Lagrangian. Even there we would have ended up with a $\max$ because of $\lambda_1, \lambda_2$ and invoking KKT.

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I think that Shiv was in the right direction. Actually, the three points he found satisfy the KKT conditions. If the lagrangean is $$ \mathcal{L} = -x_1^{1/3} - x_2 + \lambda(-I+p_1x_1 + (p_2+p_3)x_2) $$ and the KKT conditions for a minimum are $$ \frac{\partial \mathcal{L}}{\partial x_1} = -\frac13 x_1^{-2/3} + p_1 \lambda \geq 0, x_1\geq 0, \frac{\partial \mathcal{L}}{\partial x_1} x_1 = 0 $$ $$ \frac{\partial \mathcal{L}}{\partial x_2} = -1 + (p_2 + p_3) \lambda \geq 0, x_2\geq 0, \frac{\partial \mathcal{L}}{\partial x_2} x_2 = 0 $$ $$ \frac{\partial \mathcal{L}}{\partial \lambda} =-I + p_1x_1 + (p_2+p_3)x_2 \leq 0, \lambda\geq 0, \frac{\partial \mathcal{L}}{\partial \lambda} \lambda = 0 $$

From the three cases we end up with three points $$ (x_1,x_2) = \left(\left(\frac{3p_1}{p_2+p_3}\right)^{2/3}, I-3^{2/3} \left(\frac{p_1}{p_2+p_3} \right)^{5/3} \right) $$ $$ (x_1,x_2) = \left(0, \frac{I}{p_2+p_3} \right) $$ $$ (x_1,x_2) = \left(\frac{I}{p_1}, 0 \right) $$

But since the agent enjoys $x_2$ more than $x_1$ as the latter is cubic rooted, the minimum (or the maximum of the original function) will be choosing the highest $x_2$. Hence, the second point is the global minimum. We can assert this is a global minimum because of Weierstrass Theorem. The feasible region is a closed and bounded set, then the utility function (or the negative of) will achieve a maximum (minimum) value over this set.

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I try to use substitution instead of relying on KKT or the Lagrangian method until I encounter a problem that is better solved through them. In a sense, it is like people preferring different methods like elimination or substitution for solving a simple system of equations, but here we are talking about optimization problems instead of systems of equations

We can solve the problem you provided using substitution and simplifying the optimization problem.

Our UMP in consideration is: $$\begin{aligned} &\max_{x_1,x_2,x_3\geq0} \quad  x_1^\frac{1}{3} + \min(x_2,x_3)\\ &\textrm{s.t.} \quad  p_1x_1+p_2x_2+p_3x_3\leq I \end{aligned}$$

Now, what I can do is first solve the UMP holding $x_1$ constant to get demands for $x_2, x_3$ as functions of $x_1$ and other parameters: $$\begin{aligned} & \max_{x_1}  {\begin{aligned} \begin{pmatrix} \displaystyle\max_{x_2,x_3} & x_1^\frac{1}{3}+\min(x_2,x_3) \\ \textrm{s.t.}  & p_2x_2+p_3x_3\leq I-p_1x_1 \end{pmatrix} \end{aligned}}\\ \end{aligned}$$ Notice that UMP involving $x_2,x_3$ is a standard problem over perfect complements. The problem inside the brackets gives us: $$x_2(x_1,p,I)=x_3(x_1,p,I)=\frac{I-p_1x_1}{p_2+p_3}$$

We can now use the above, substitute them into the problem inside the bracket of the original UMP and solve the optimization to get optimal $x_1^*$: $$\begin{aligned} &\max_{0 \leq x_1\leq\frac{I}{p_1}} \quad x_1^\frac{1}{3}+\frac{I-p_1x_1}{p_2+p_3}\\ \text{gives} \quad & x_1^*(p,I)=\left({\frac{p_2+p_3}{3p_1}}\right)^\frac{3}{2} \end{aligned}$$

Substituting $x_1^*$ in the expression for $x_2,x_3$ that we derived before gives us the final demands:

$$(x_1^*,x_2^*,x_3^*)(p,I)=\begin{cases} \left (\left({\frac{p_2+p_3}{3p_1}}\right)^\frac{3}{2},\frac{I}{p_2+p_3}-\frac{1}{3}\left(\frac{p_2+p_3}{3p_1}\right)^\frac{1}{2}, \frac{I}{p_2+p_3}-\frac{1}{3}\left(\frac{p_2+p_3}{3p_1}\right)^\frac{1}{2}\right) & \text{if } I\sqrt{p_1} > \left(\frac{p_2+p_3}{3}\right)^\frac{3}{2}\\ \left (\left({\frac{p_2+p_3}{3p_1}}\right)^\frac{3}{2},0, 0 \right) & \text{if } I\sqrt{p_1} \leq \left(\frac{p_2+p_3}{3}\right)^\frac{3}{2} \end{cases}$$

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