Would this hypothetical firm exit in the long run?

Question

I'm trying to work out the following problem for my microeconomics course, using the standard method we have been taught.

Problem: Output is a function of inputs $x_1$ and $x_2$, $y=x_1^{1/2}x_2$. Total cost is a function of inputs $x_1$ and $x_2$ and their corresponding prices, $w_1=3$ and $w_2=2$, $c=3x_1+2x_2$. The price at which output can be sold is $p=6$. What is the profit maximizing level of $x_2$ for the firm to use in the long run?

Attempt: In the long run, the profit maximizing level of $x_2$ is found by solving $p\times MP_2=w_2$. That is, $6x_1^{1/2}=2$. But, $x_2$ is no longer a variable in this equation! This leads me to believe the firm will exit, meaning $x_2=0$, but I cannot think of how to relate the economic theory to the mathematics.

Answer 1

The firm will not exit: for example inputs $x_1=1,x_2=1$ lead to $y=1, c=5$ and profit $py-c=1$

In general profit is $6x_1^{1/2}x_2-3x_1-2x_2$

• As you found with $x_1 =\frac19$, profit is $-\frac13$ which is independent of $x_2$

• But with $x_1 \gt \frac19$, profit is an increasing function of $x_2$ and is positive when $x_2 \gt \frac{3x_1}{6\sqrt{x_1}-2}$

• For example with $x_1 =1$, output is $x_2$, profit is $4x_2 -3$ and is positive for $x_2 \gt \frac34$

• For example with $x_1 =4$, output is $2x_2$, profit is $10x_2 -12$ and is positive for $x_2 \gt \frac{6}{5}$

So there appears to be no profit maximizing level of $x_2$ because (providing $x_1 \gt \frac19$) profit is an increasing function of $x_2$. So a profit maximiser will want to increase $x_2$ without limit. Your partial derivative method did not find this infinite solution

For what it is worth, given a particular $x_2$, the optimal value of $x_1$ seems to be $x_2^2$ making the profit $x_2(3x_2-2)$ which is clearly positive when $x_2>\frac23$ and then again an increasing function of $x_2$