# Cost Minimization and Karush-Kuhn-Tucker

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: $$$$\mathcal{L} = w_1 x_1 + w_2 x_2 + \lambda_1 (\sqrt{x_1 x_2} - \bar{y}) + \lambda_2 (x_1 -1)$$$$ But when I consider the first order conditions, I obtained that $$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.

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, $$C(\omega_1,\omega_2,\bar{y})= \begin{cases}\omega_1+\omega_2\bar{y}^2, & \frac{\omega_2}{\omega_1}(\bar{y}^2) \le 1 \\ 2(\omega_1\omega_2)^\frac{1}{2}\bar{y},& \text{otherwise} \end{cases}$$

• Cost function is not equal to $\min\{\omega_1+\omega_2(\overline{y})^2, 2(\omega_1\omega_2)^\frac{1}{2}\overline{y} \}$
– Amit
Commented May 21 at 11:50
• 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 $2(w_1w_2)^{\frac{1}{2}}\bar{y}=1$. Commented May 21 at 13:17
• Thanks for your inputs, I made the necessary changes to my answer.
– SGP
Commented May 22 at 6:13