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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.

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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 $x_2$

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