A firm produces an output $y$ using two inputs $x_1$ and $x_2$, where the production function is given by $y = \sqrt{x_1 x_2}$ for any $(x_1, x_2) \in \mathbb{R}^2_+$. Union agreements obligate the firm to use at least one unit of $x_1$ in its production process. The input prices of $x_1$ and $x_2$ are given by $w_1$ and $w_2$, respectively. The firm is supposed to produce $\bar{y}$ units of output due to its commitments. I want to find the minimum cost. When I set up the Lagrangian I obtained: \begin{equation} \mathcal{L} = w_1 x_1 + w_2 x_2 + \lambda_1 (\sqrt{x_1 x_2} - \bar{y}) + \lambda_2 (x_1 -1) \end{equation} But when I consider the first order conditions, I obtained that $x_1 = 1$, but this does not seem reasonable to me. I am wondering if there is a mistake in my Lagrangian and first order condition. Any help is appreciated.


3 Answers 3


The term $\lambda_2(x_1-1)$ in your Lagrangian is incorrect; it treats the second constraint as an equality rather than an inequality. To allow for the constraint being an inequality you can include a slack variable, say $a$, and rewrite the term as $\lambda_2(x_1-1-a^2)$. You can then proceed to calculate the first-order conditions, but these will now need to include: $\partial \mathcal{L}/\partial a=0$. Note that the first-order condition:

$$\frac{\partial \mathcal{L}}{\partial \lambda_2}=x_1-1-a^2=0$$

implies $x_1\geq1$ since $a^2$ must be non-negative.


Here is the cost minimisation problem that we need to solve: \begin{eqnarray*} \min_{x_1,x_2} & w_1x_1+w_2x_2 \\ \text{s.t. } & \sqrt{x_1x_2}=\overline{y} \\ \text{and } & x_1\geq 1, x_2\geq 0 \end{eqnarray*} where $w_1>0$, $w_2>0$, $\overline{y}>0$ are given. Solving this problem, we get the conditional input demands as follows: \begin{eqnarray*}(x_1^c,x_2^c)(w_1,w_2,\overline{y})=\begin{cases}\left(\sqrt{\dfrac{w_2\overline{y}^2}{w_1}},\sqrt{\dfrac{w_1\overline{y}^2}{w_2}}\right) & \text{if } \dfrac{w_2\overline{y}^2}{w_1}\geq 1 \\ \left(1,\overline{y}^2\right) & \text{if } \dfrac{w_2\overline{y}^2}{w_1}< 1\end{cases}\end{eqnarray*} and the cost function is: \begin{eqnarray*} C(w_1,w_2,\overline{y})=\begin{cases} (2\sqrt{w_1w_2})\overline{y} & \text{if } \dfrac{w_2\overline{y}^2}{w_1}\geq 1 \\ w_1+w_2\overline{y}^2 & \text{if } \dfrac{w_2\overline{y}^2}{w_1}< 1\end{cases}\end{eqnarray*}


Let's setup the optimization problem first, $$min_{\{x_1,x_2\}} \omega_1x_1+\omega_2x_2 $$ $$ s.t \hspace{5 mm} (\bar{y}=x_1^{\frac{1}{2}}x_2^{\frac{1}{2}}) \wedge(x_1 \ge 1)\wedge (x_2 \ge 0)$$

Discussion before solving the optimization problem, here we have the objective function as convex, which means that the Local Minimizer is a Global Minimizer.

Now lets consider the Lagrangian, $$\mathcal{L}(x_1,x_2)= \omega_1x_1+\omega_2x_2-\lambda(x_1^{\frac{1}{2}}x_2^{\frac{1}{2}}-\bar{y})-\mu_{1}(x_1-1)-\mu_2x_2$$ Now as per the KKT first order conditions, we have,

$$\frac{\partial \mathcal{L}}{\partial x_1}=\omega_1-(\frac{\lambda}{2}x_1^{\frac{-1}{2}}x_2^{\frac{1}{2}})-\mu_1=0$$ $$\frac{\partial \mathcal{L}}{\partial x_2}=\omega_2-(\frac{\lambda}{2}x_1^{\frac{1}{2}}x_2^{\frac{-1}{2}})-\mu_2=0$$ $$\text{and} \hspace{2mm} \bar{y}= x_1^{\frac{1}{2}}x_2^{\frac{1}{2}}; \hspace{2mm}\lambda\ge0;\hspace{2mm} \mu_1\ge0;\hspace{2mm} \mu_2\ge0 $$ $$ \lambda(\bar{y}-x_1^{\frac{1}{2}}x_2^{\frac{1}{2}})=0;\hspace{2mm}\mu_{1}(x_1-1)=0;\hspace{2mm}\mu_2x_2=0$$

After some mind-numbing algebra, we get

case 1: $$x_1=1, x_2=\bar{y}^2, \lambda= 2\omega_2 \bar{y}, \mu_1=-\omega_2(\bar{y}^2)+\omega_1,\mu_2=0$$

Here as $\mu_1 \ge 0$ $\iff \frac{\omega_2}{\omega_1}(\bar{y}^2) \le 1 $

and case 2: $$x_1=(\frac{\omega_2}{\omega_1})^\frac{1}{2}\bar{y}, x_2=(\frac{\omega_1}{\omega_2})^\frac{1}{2}\bar{y}, \lambda=2(\omega_1\omega_2)^\frac{1}{2},\mu_1=0,\mu_2=0$$

Therefore, we will end up with the conditional factor demands, $$(x_1^c,x_2^c)=\begin{cases}(1,\bar{y}^2), & \frac{\omega_2}{\omega_1}(\bar{y}^2) \le 1 \\ ((\frac{\omega_2}{\omega_1})^\frac{1}{2}\bar{y},(\frac{\omega_1}{\omega_2})^\frac{1}{2}\bar{y}),& \text{otherwise} \end{cases}$$ Hence we will get the cost function as, $$C(\omega_1,\omega_2,\bar{y})= \begin{cases}\omega_1+\omega_2\bar{y}^2, & \frac{\omega_2}{\omega_1}(\bar{y}^2) \le 1 \\ 2(\omega_1\omega_2)^\frac{1}{2}\bar{y},& \text{otherwise} \end{cases}$$

  • 1
    $\begingroup$ Cost function is not equal to $\min\{\omega_1+\omega_2(\overline{y})^2, 2(\omega_1\omega_2)^\frac{1}{2}\overline{y} \}$ $\endgroup$
    – Amit
    Commented May 21 at 11:50
  • 1
    $\begingroup$ For example, if $w_1=w_2=1$ and $\bar{y}=\frac{1}{2}$, the minimum cost solution meeting the constraint is $(x_1,x_2)=(1,\frac{1}{4})$ with cost $1.25$ but $2(w_1w_2)^{\frac{1}{2}}\bar{y}=1$. $\endgroup$ Commented May 21 at 13:17
  • $\begingroup$ Thanks for your inputs, I made the necessary changes to my answer. $\endgroup$
    – SGP
    Commented May 22 at 6:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.