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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 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 at 22:11
  • $\begingroup$ "by noting that for the utility to be the maximum." - something seems to be missing here. $\endgroup$ – VARulle Jun 14 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 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 at 3:00
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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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