<p>Similar question to what you ask is: <em><strong>How can I formally solve the expenditure minimization problem (EMP) using the Kuhn-Tucker conditions, given a Cobb-Douglas utility function <span class="math-container">$u(x_1,x_2)=x_1^\alpha x_2^{1-\alpha}$</span>?</strong></em></p>
<hr />
<p>The expenditure minimization problem (EMP) for a Cobb-Douglas utility function <span class="math-container">$u(x_1,x_2) = x_1^{\alpha}x_2^{1-\alpha}$</span> subject to a budget constraint can be formulated as follows:</p>
<p><span class="math-container">\begin{aligned}
\text{minimize } &amp; p_1x_1 + p_2x_2 \\
\text{subject to } &amp; u(x_1,x_2) \geq u \\
&amp; p_1x_1 + p_2x_2 \leq w \\
&amp; x_1 \geq 0, x_2 \geq 0
\end{aligned}</span></p>
<p>where <span class="math-container">$p_1$</span> and <span class="math-container">$p_2$</span> are the prices of goods 1 and 2, <span class="math-container">$w$</span> is the budget, and <span class="math-container">$\alpha \in (0,1)$</span>.</p>
<p>The Lagrangian for this problem is:</p>
<p><span class="math-container">$L(x_1,x_2,\lambda) = p_1x_1 + p_2x_2 - \lambda(u(x_1,x_2) - u)$</span></p>
<p>The vector form of <span class="math-container">$\nabla_x L$</span> is:</p>
<p><span class="math-container">$$\nabla_x L = \begin{bmatrix} \frac{\partial L}{\partial x_1} \ \frac{\partial L}{\partial x_2} \ \frac{\partial L}{\partial \lambda} \end{bmatrix} = \begin{bmatrix} p_1 - \lambda \alpha x_1^{\alpha-1}x_2^{1-\alpha} \ p_2 - \lambda (1-\alpha) x_1^{\alpha}x_2^{-\alpha} \ u(x_1,x_2) - u \end{bmatrix}$$</span></p>
<p>So the <a href="https://en.wikipedia.org/wiki/Karush%E2%80%93Kuhn%E2%80%93Tucker_conditions" rel="nofollow noreferrer">Kuhn-Tucker conditions</a> for a minimum can be written as:</p>
<p><span class="math-container">$$\nabla_x L = \begin{bmatrix} p_1 - \lambda \alpha x_1^{\alpha-1}x_2^{1-\alpha} &amp; p_2 - \lambda (1-\alpha) x_1^{\alpha}x_2^{-\alpha} &amp; u(x_1,x_2) - u \end{bmatrix} = \mathbf{0}$$</span></p>
<p>The Kuhn-Tucker conditions for a minimum are as well:</p>
<ul>
<li><span class="math-container">$\lambda(u(x_1,x_2) - u) = 0$</span></li>
<li><span class="math-container">$u(x_1,x_2) \geq u$</span></li>
<li><span class="math-container">$p_1x_1 + p_2x_2 \leq w$</span></li>
<li><span class="math-container">$x_1 \geq 0, x_2 \geq 0$</span></li>
<li><span class="math-container">$\lambda \geq 0$</span></li>
</ul>
<p>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.</p>
<p>The Hessian matrix <span class="math-container">$H(x_1,x_2,\lambda)$</span> is a 3x3 matrix and is given by:</p>
<p><span class="math-container">$$H(x_1,x_2,\lambda) = \begin{bmatrix} \frac{\partial^2 L}{\partial x_1^2} &amp; \frac{\partial^2 L}{\partial x_1\partial x_2} &amp; \frac{\partial^2 L}{\partial x_1\partial \lambda} \\ \frac{\partial^2 L}{\partial x_2\partial x_1} &amp; \frac{\partial^2 L}{\partial x_2^2} &amp; \frac{\partial^2 L}{\partial x_2\partial \lambda} \\ \frac{\partial^2 L}{\partial \lambda\partial x_1} &amp; \frac{\partial^2 L}{\partial \lambda\partial x_2} &amp; \frac{\partial^2 L}{\partial \lambda^2} \end{bmatrix}$$</span></p>
<p>If we evaluate the Hessian at the solution <span class="math-container">$$x_1 = u_{x_1}, x_2 = u_{x_2}, \lambda = \frac{1}{u_{x_1}^\alpha u_{x_2}^{1-\alpha}}$$</span> we get:</p>
<p><span class="math-container">$$H(x_1,x_2,\lambda) = \begin{bmatrix} 0 &amp; -\alpha \lambda x_2^{1-\alpha}x_1^{\alpha-1} &amp; -\alpha \lambda x_2^{1-\alpha}x_1^{\alpha-1} \\ -\alpha \lambda x_2^{1-\alpha}x_1^{\alpha-1} &amp; 0 &amp; (1-\alpha)\lambda x_1^{\alpha}x_2^{-\alpha} \\ -\alpha \lambda x_2^{1-\alpha}x_1^{\alpha-1} &amp; (1-\alpha)\lambda x_1^{\alpha}x_2^{-\alpha} &amp; 0 \end{bmatrix}$$</span></p>
<p>The determinant of the Hessian is:</p>
<p><span class="math-container">$$\det(H) = 2\alpha^2(1-\alpha)^2\lambda^3 x_1^{2\alpha-2}x_2^{-2\alpha-1}$$</span></p>
<p>At the optimal point, we have <span class="math-container">$\alpha x_1^{\alpha}x_2^{1-\alpha} = u$</span> and <span class="math-container">$\lambda &gt; 0$</span>, so we can substitute these expressions into the determinant and get:</p>
<p><span class="math-container">$$\det(H) = 2\alpha^2(1-\alpha)^2\lambda^3\frac{u}{x_1^2x_2} &gt; 0$$</span></p>
<p>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.</p>

CC BY-SA 4.0
2023-02-20T09:17:33.220Z
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
Dimitru
CC BY-SA 4.0
2023-02-19T11:25:38.290Z
2023-02-21T12:06:40.663Z
2023-02-21T12:06:40.663Z
0x90
<p>How should I formally solve the expenditure min.problem (EMP) by using KT conditions?</p>
<p>Since I should follow the notation of the Mas-Colell, I should write:</p>
<p><span class="math-container">$\min~$</span> <span class="math-container">$p \cdot x$</span> , s.t. <span class="math-container">$u(x) \ge u$</span></p>
<p>Which shall be restated as <span class="math-container">$\max \ -p \cdot x$</span>  , s.t. <span class="math-container">$-u(x) \le -u$</span></p>
<p>And the first order conditions are:</p>
<p><span class="math-container">$p_l \ge \lambda \frac{\partial u(x^{\ast})}{x_l}$</span>, with equality for <span class="math-container">$x_l^{\ast}&gt;0$</span>, for all <span class="math-container">$l=1,2,...,L$</span></p>
<p>Suppose we have a Cobb-Douglas utility function <span class="math-container">$u(x_1,x_2)=x_1^\alpha x_2^{1-\alpha}$</span></p>
<p>The minimization problem yields:</p>
<p><span class="math-container">$\max \ -(p_1 x_1 + p_2 x_2)$</span>, <span class="math-container">$~~s.t. ~~~-x_1^\alpha x_2^{1-\alpha} \le -u $</span></p>
<p>Preferences are strongly monotone, which implies an interior solution for <span class="math-container">$u&gt;0$</span>.</p>
<p>How can formally state that the inequality constraint holds as equality constraint ?</p>
<p>Also, how can I formally state that what I find is a minimum?</p>

<microeconomics><utility><optimization>
Kuhn-Tucker(KT) conditons EMP
Dimitru
# Kuhn-Tucker(KT) conditons EMP

(Asked by: username_0 on 2023-02-17)

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?

**Tags:** microeconomics, utility, optimization


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## 1 answers

### Answer by username_1 on 2023-02-19. Score: 1

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 $u(x\_1,x\_2)=x\_1^\alpha x\_2^{1-\alpha}$?***

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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](https://en.wikipedia.org/wiki/Karush%E2%80%93Kuhn%E2%80%93Tucker_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.

> **Comments:**
> **username_0**: 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  

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