# Would this hypothetical firm exit in the long run?

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.

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 $$x_2$$