Given $f(x_1,x_2) = \min\{x_1,x_2\}^\alpha$ find the profit-maximizing factor demands, supply function and profit fuction

Question

Given

$$f(x_1,x_2) = \min\{x_1,x_2\}^\alpha$$

I have to find the profit-maximizing demand functions, supply function and profit function but I am not sure how to when the function is given as it is.

I know that I for example have to use Lagrange. Thus

$$ L = p \cdot f(x_1,x_2) - w_1 \cdot x_1 - w_2 \cdot x_2 - \lambda \left(p_1x_1 + p_2x_2 - m \right) $$ and then afterwards find the first order conditions. But how I find the derivative in regards to both $x_1$ and $x_2$ when I have a function such as $f(x_1,x_2) = \min\{x_1,x_2\}^\alpha$ ?

Can you help me in the right direction?

Thanks in advance!

Answer 1

Observe that if $x_1^{*} > x_2^{*}$, then $$\pi(x_1^{*}, x_2^{*}) = p {x_2^*}^{\alpha} - w_1x_1^{*} - w_2x_2^* < p {x_2^*}^{\alpha} - w_1x_2^{*} - w_2x_2^* = \pi(x_2^*, x_2^*)$$

Due to symmetry, we can conclude that the optimal $x_1$ and $x_2$ will be equal.

From the profit function $\pi(w_1, w_2) = \max_{x \geq 0} [px^\alpha - w_1x - w_2x]$, we get

$$x_i^*(p,w_1, w_2) = \left(\frac{p \alpha}{w_1 + w_2}\right)^{\frac{1}{1-\alpha}}$$ for $i=1,2$

Calculate $\Pi^*(p,w_1,w_2)$ using the factor demands we found above.

Answer 2

Given the production function $f:\mathbb{R}^2_+\rightarrow\mathbb{R}$ defined as $f(x_1,x_2) = (\min(x_1,x_2))^\alpha$, where $\alpha > 0$, we can solve the profit maximisation problem of the competitive firm in two steps.

In step 1, we solve the cost minimisation problem: \begin{eqnarray*}\min_{x_1\geq 0, x_2\geq 0} & w_1x_1+w_2x_2\\ \text{s.t. } & (\min(x_1,x_2))^\alpha \geq y\end{eqnarray*} where $w_1 > 0, w_2 > 0$ and $y \geq 0$.

Solving the problem gives conditional input demands as: \begin{eqnarray*}(x_1^c,x_2^c)(w_1,w_2,y) = (y^{\frac{1}{\alpha}}, y^{\frac{1}{\alpha}})\end{eqnarray*}

and the cost function is \begin{eqnarray*}c(w_1,w_2,y) = (w_1+w_2)y^{\frac{1}{\alpha}}\end{eqnarray*}

In step 2, we'll solve the profit maximisation problem which is \begin{eqnarray*}\max_{y\geq 0} & \ py - (w_1+w_2)y^{\frac{1}{\alpha}}\end{eqnarray*} where $p>0$, $w_1>0$ and $w_2>0$

Here we'll consider three cases for $\alpha$:

• $0 < \alpha < 1$ (Case of Decreasing returns to scale): Solving the profit maximisation problem in this case will give the supply function as: \begin{eqnarray*}y^s(w_1,w_2,p) = \left(\frac{\alpha p}{ w_1+w_2}\right)^{\frac{\alpha}{1-\alpha}}\end{eqnarray*} and the input demands are \begin{eqnarray*}(x_1^d,x_2^d)(w_1,w_2,p) = \left(\left(\frac{\alpha p}{ w_1+w_2}\right)^{\frac{1}{1-\alpha}}, \left(\frac{\alpha p}{ w_1+w_2}\right)^{\frac{1}{1-\alpha}}\right)\end{eqnarray*} Profit function is \begin{eqnarray*}\pi(w_1,w_2,p) = p\left(\frac{\alpha p}{ w_1+w_2}\right)^{\frac{\alpha}{1-\alpha}} - (w_1+w_2)\left(\frac{\alpha p}{ w_1+w_2}\right)^{\frac{1}{1-\alpha}}\end{eqnarray*}

• $\alpha = 1$ (Case of Constant returns to scale): In this case, supply is \begin{eqnarray*}y^s(w_1,w_2,p) \in \begin{cases} \{0\} & \text{if } p < w_1+w_2 \\ \mathbb{R}_+ & \text{if } p = w_1+w_2 \\ \emptyset & \text{if } p > w_1+w_2 \end{cases} \end{eqnarray*} and the input demands are \begin{eqnarray*}(x_1^d,x_2^d)(w_1,w_2,p) = (x_1^c,x_2^c)(w_1,w_2,y^s(w_1,w_2,p))\end{eqnarray*} Here, profit function is only defined for the case where $p\leq w_1+w_2$ and is given by \begin{eqnarray*}\pi(w_1,w_2,p) = 0\end{eqnarray*}

• $\alpha > 1$ (Case of Increasing returns to scale): Here the solution to the profit maximisation problem does not exist i.e. supply does not exist.