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?
Comment(s): I am unsure on how to set up my Lagrange multiplier function. I know the general function form, but this is a type of problem I have not yet encountered. Moreover, I am unsure of how to set up my constraint. (I am very much used to have a constraint function just given to me) any help is appreciated.
EDIT 1:
$\frac{\partial \mathcal{L}}{\partial x_1}=-\frac{1}{3}x^{2/3}-\lambda+(\lambda p\cdot x_1)=0$
$\frac{\partial \mathcal{L}}{\partial x_2}= -1-\lambda+\lambda(p_2+p_3)x_2=0$
$\frac{\partial \mathcal{L}}{\partial \lambda}=-x_1-x_2+\lambda(p_1x_1+(p_2+p_3)x_2-I)=0$
$p_2=p_3$
$\lambda = 1$
$x_2=1$
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.