Q. 3 A consumer consumes two goods, and the utility from consuming an amount \( x_{i} \) of commodity \( i=1,2 \) is given by \( u\left(x_{1}, x_{2}\right)=x_{1} x_{2}^{2} \). The consumer has an income \( I>0 \), and the price of commodity \( i \) is \( p_{i}>0 \). We maximize the utility \( u \) under the constraints \[ I-p_{1} x_{1}-p_{2} x_{2}=0, x_{1}>0, x_{2}>0 . \] (1) Find the Lagrangian of this maximization problem with constraints. (2) Find the critical points of the Lagrangian. (3) Solve the maximization problem.

**Problem 3** A consumer aims to maximize their utility given by \( u(x_1, x_2) = x_1 x_2^2 \) subject to their budget constraint \( p_1 x_1 + p_2 x_2 = I \), where \( I > 0 \) is the income and \( p_1, p_2 > 0 \) are the prices of commodities 1 and 2, respectively. --- ### (1) Lagrangian of the Maximization Problem To incorporate the budget constraint into the utility maximization problem, we use the method of Lagrange multipliers. The Lagrangian \( \mathcal{L} \) is formulated as: \[ \mathcal{L}(x_1, x_2, \lambda) = u(x_1, x_2) + \lambda \left( I - p_1 x_1 - p_2 x_2 \right) \] Substituting the given utility function: \[ \mathcal{L}(x_1, x_2, \lambda) = x_1 x_2^2 + \lambda \left( I - p_1 x_1 - p_2 x_2 \right) \] --- ### (2) Critical Points of the Lagrangian To find the critical points, we take the partial derivatives of the Lagrangian with respect to \( x_1 \), \( x_2 \), and \( \lambda \) and set them to zero. 1. **Partial derivative with respect to \( x_1 \):** \[ \frac{\partial \mathcal{L}}{\partial x_1} = x_2^2 - \lambda p_1 = 0 \] \[ x_2^2 = \lambda p_1 \quad \text{(1)} \] 2. **Partial derivative with respect to \( x_2 \):** \[ \frac{\partial \mathcal{L}}{\partial x_2} = 2x_1 x_2 - \lambda p_2 = 0 \] \[ 2x_1 x_2 = \lambda p_2 \quad \text{(2)} \] 3. **Partial derivative with respect to \( \lambda \):** \[ \frac{\partial \mathcal{L}}{\partial \lambda} = I - p_1 x_1 - p_2 x_2 = 0 \] \[ I = p_1 x_1 + p_2 x_2 \quad \text{(3)} \] These equations (1), (2), and (3) constitute the system of equations for the critical points. --- ### (3) Solving the Maximization Problem To solve for \( x_1 \) and \( x_2 \), we can eliminate \( \lambda \) from equations (1) and (2). From equation (1): \[ \lambda = \frac{x_2^2}{p_1} \] Substitute \( \lambda \) into equation (2): \[ 2x_1 x_2 = \left( \frac{x_2^2}{p_1} \right) p_2 \] \[ 2x_1 x_2 = \frac{p_2}{p_1} x_2^2 \] \[ 2x_1 = \frac{p_2}{p_1} x_2 \] \[ x_1 = \frac{p_2}{2 p_1} x_2 \quad \text{(4)} \] Now, substitute equation (4) into the budget constraint (3): \[ I = p_1 \left( \frac{p_2}{2 p_1} x_2 \right) + p_2 x_2 \] \[ I = \frac{p_2}{2} x_2 + p_2 x_2 \] \[ I = \frac{3 p_2}{2} x_2 \] \[ x_2 = \frac{2I}{3 p_2} \] Now, substitute \( x_2 \) back into equation (4) to find \( x_1 \): \[ x_1 = \frac{p_2}{2 p_1} \cdot \frac{2I}{3 p_2} \] \[ x_1 = \frac{I}{3 p_1} \] **Solution:** \[ \boxed{ \begin{aligned} x_1^* &= \frac{I}{3 p_1} \\ x_2^* &= \frac{2I}{3 p_2} \end{aligned} } \] These values \( x_1^* \) and \( x_2^* \) maximize the consumer's utility subject to the given budget constraint. --- **Summary of Answers:** 1. **Lagrangian:** \[ \mathcal{L} = x_1 x_2^2 + \lambda (I - p_1 x_1 - p_2 x_2) \] 2. **Critical Points:** \[ \begin{cases} x_2^2 = \lambda p_1 \\ 2x_1 x_2 = \lambda p_2 \\ I = p_1 x_1 + p_2 x_2 \end{cases} \] 3. **Optimal Consumption Bundle:** \[ x_1^* = \frac{I}{3 p_1}, \quad x_2^* = \frac{2I}{3 p_2} \]