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.
