<p>The firm will not exit: for example inputs <span class="math-container">$x_1=1,x_2=1$</span> lead to <span class="math-container">$y=1, c=5$</span> and profit <span class="math-container">$py-c=1$</span></p>

<p>In general profit is <span class="math-container">$6x_1^{1/2}x_2-3x_1-2x_2$</span>  </p>

<ul>
<li><p>As you found with <span class="math-container">$x_1 =\frac19$</span>, profit is <span class="math-container">$-\frac13$</span> which is independent of <span class="math-container">$x_2$</span></p></li>
<li><p>But with <span class="math-container">$x_1 \gt \frac19$</span>, profit is an increasing function of <span class="math-container">$x_2$</span> and is positive when <span class="math-container">$x_2 \gt \frac{3x_1}{6\sqrt{x_1}-2}$</span></p></li>
<li><p>For example with <span class="math-container">$x_1 =1$</span>, output is <span class="math-container">$x_2$</span>, profit is <span class="math-container">$4x_2 -3$</span> and is positive for <span class="math-container">$x_2 \gt \frac34$</span></p></li>
<li>For example with <span class="math-container">$x_1 =4$</span>, output is <span class="math-container">$2x_2$</span>, profit is <span class="math-container">$10x_2 -12$</span> and is positive for <span class="math-container">$x_2 \gt \frac{6}{5}$</span></li>
</ul>

<p>So there appears to be no  profit maximizing level of <span class="math-container">$x_2$</span> because (providing <span class="math-container">$x_1 \gt \frac19$</span>) profit is an increasing  function of <span class="math-container">$x_2$</span>. So a profit maximiser will want to increase <span class="math-container">$x_2$</span> without limit.  Your partial derivative method did not find this infinite solution</p>

<p>For what it is worth, given a particular <span class="math-container">$x_2$</span>, the optimal value of <span class="math-container">$x_1$</span> seems to be <span class="math-container">$x_2^2$</span> making the profit <span class="math-container">$x_2(3x_2-2)$</span> which is clearly positive when <span class="math-container">$x_2&gt;\frac23$</span> and then again an increasing function of <span class="math-container">$x_2$</span></p>

CC BY-SA 4.0
2018-12-10T09:10:34.657Z
2018-12-10T09:10:34.657Z
Henry
<p>I'm trying to work out the following problem for my microeconomics course, using the standard method we have been taught.</p>

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

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

<microeconomics>
Would this hypothetical firm exit in the long run?
Karambwan
# Would this hypothetical firm exit in the long run?

(Asked by: username_0 on 2018-12-10)

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.

**Tags:** microeconomics

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## 1 answers

### Answer by username_1 on 2018-12-10. Score: 0 | Accepted answer

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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