# Kuhn-Tucker(KT) conditons EMP

How should I formally solve the expenditure min.problem (EMP) by using KT conditions?

Since I should follow the notation of the Mas-Colell, I should write:

$$\min~$$ $$p \cdot x$$ , s.t. $$u(x) \ge u$$

Which shall be restated as $$\max \ -p \cdot x$$ , s.t. $$-u(x) \le -u$$

And the first order conditions are:

$$p_l \ge \lambda \frac{\partial u(x^{\ast})}{x_l}$$, with equality for $$x_l^{\ast}>0$$, for all $$l=1,2,...,L$$

Suppose we have a Cobb-Douglas utility function $$u(x_1,x_2)=x_1^\alpha x_2^{1-\alpha}$$

The minimization problem yields:

$$\max \ -(p_1 x_1 + p_2 x_2)$$, $$~~s.t. ~~~-x_1^\alpha x_2^{1-\alpha} \le -u$$

Preferences are strongly monotone, which implies an interior solution for $$u>0$$.

How can formally state that the inequality constraint holds as equality constraint ?

Also, how can I formally state that what I find is a minimum?

Similar question to what you ask is: How can I formally solve the expenditure minimization problem (EMP) using the Kuhn-Tucker conditions, given a Cobb-Douglas utility function $$u(x_1,x_2)=x_1^\alpha x_2^{1-\alpha}$$?

The expenditure minimization problem (EMP) for a Cobb-Douglas utility function $$u(x_1,x_2) = x_1^{\alpha}x_2^{1-\alpha}$$ subject to a budget constraint can be formulated as follows:

\begin{aligned} \text{minimize } & p_1x_1 + p_2x_2 \\ \text{subject to } & u(x_1,x_2) \geq u \\ & p_1x_1 + p_2x_2 \leq w \\ & x_1 \geq 0, x_2 \geq 0 \end{aligned}

where $$p_1$$ and $$p_2$$ are the prices of goods 1 and 2, $$w$$ is the budget, and $$\alpha \in (0,1)$$.

The Lagrangian for this problem is:

$$L(x_1,x_2,\lambda) = p_1x_1 + p_2x_2 - \lambda(u(x_1,x_2) - u)$$

The vector form of $$\nabla_x L$$ is:

$$\nabla_x L = \begin{bmatrix} \frac{\partial L}{\partial x_1} \ \frac{\partial L}{\partial x_2} \ \frac{\partial L}{\partial \lambda} \end{bmatrix} = \begin{bmatrix} p_1 - \lambda \alpha x_1^{\alpha-1}x_2^{1-\alpha} \ p_2 - \lambda (1-\alpha) x_1^{\alpha}x_2^{-\alpha} \ u(x_1,x_2) - u \end{bmatrix}$$

So the Kuhn-Tucker conditions for a minimum can be written as:

$$\nabla_x L = \begin{bmatrix} p_1 - \lambda \alpha x_1^{\alpha-1}x_2^{1-\alpha} & p_2 - \lambda (1-\alpha) x_1^{\alpha}x_2^{-\alpha} & u(x_1,x_2) - u \end{bmatrix} = \mathbf{0}$$

The Kuhn-Tucker conditions for a minimum are as well:

• $$\lambda(u(x_1,x_2) - u) = 0$$
• $$u(x_1,x_2) \geq u$$
• $$p_1x_1 + p_2x_2 \leq w$$
• $$x_1 \geq 0, x_2 \geq 0$$
• $$\lambda \geq 0$$

To show that the solution obtained from the Kuhn-Tucker conditions is a minimum, we need to verify that the Hessian of the Lagrangian is positive definite at the solution.

The Hessian matrix $$H(x_1,x_2,\lambda)$$ is a 3x3 matrix and is given by:

$$H(x_1,x_2,\lambda) = \begin{bmatrix} \frac{\partial^2 L}{\partial x_1^2} & \frac{\partial^2 L}{\partial x_1\partial x_2} & \frac{\partial^2 L}{\partial x_1\partial \lambda} \\ \frac{\partial^2 L}{\partial x_2\partial x_1} & \frac{\partial^2 L}{\partial x_2^2} & \frac{\partial^2 L}{\partial x_2\partial \lambda} \\ \frac{\partial^2 L}{\partial \lambda\partial x_1} & \frac{\partial^2 L}{\partial \lambda\partial x_2} & \frac{\partial^2 L}{\partial \lambda^2} \end{bmatrix}$$

If we evaluate the Hessian at the solution $$x_1 = u_{x_1}, x_2 = u_{x_2}, \lambda = \frac{1}{u_{x_1}^\alpha u_{x_2}^{1-\alpha}}$$ we get:

$$H(x_1,x_2,\lambda) = \begin{bmatrix} 0 & -\alpha \lambda x_2^{1-\alpha}x_1^{\alpha-1} & -\alpha \lambda x_2^{1-\alpha}x_1^{\alpha-1} \\ -\alpha \lambda x_2^{1-\alpha}x_1^{\alpha-1} & 0 & (1-\alpha)\lambda x_1^{\alpha}x_2^{-\alpha} \\ -\alpha \lambda x_2^{1-\alpha}x_1^{\alpha-1} & (1-\alpha)\lambda x_1^{\alpha}x_2^{-\alpha} & 0 \end{bmatrix}$$

The determinant of the Hessian is:

$$\det(H) = 2\alpha^2(1-\alpha)^2\lambda^3 x_1^{2\alpha-2}x_2^{-2\alpha-1}$$

At the optimal point, we have $$\alpha x_1^{\alpha}x_2^{1-\alpha} = u$$ and $$\lambda > 0$$, so we can substitute these expressions into the determinant and get:

$$\det(H) = 2\alpha^2(1-\alpha)^2\lambda^3\frac{u}{x_1^2x_2} > 0$$

Since the determinant is positive and the diagonal elements are zero, the Hessian is positive semi-definite. Therefore, the solution is a minimum if the Hessian is positive semi-definite.

• This does not answer to my question. Btw, there is an error, I've never put the maximum wealth level to spend as constraint. This is a mistake, the level of expenditure is endogenously determined. Then, you switched from Kuhn-tucker to lagrangian conditions arbitrarily. I'm surprised they upvoted your answer. It is wrong Feb 20, 2023 at 9:17