<p>We need to solve the following problem:</p>
<p><span class="math-container">$$
\min_{q_A, q_B}\left\{50 \times I_{q_A &gt; 0} + 5 q_A + q_B^2\right\} \text{ s.t. } q_A + q_B = q \text{ and } q_A, q_B \ge 0
$$</span>
We split up according to whether <span class="math-container">$q_A = 0$</span> or <span class="math-container">$q_A &gt; 0$</span>.</p>
<ul>
<li><p>If <span class="math-container">$q_A = 0$</span> then <span class="math-container">$q_B = q$</span> and we have a total cost of <span class="math-container">$q^2$</span>.</p>
</li>
<li><p>If <span class="math-container">$q_A &gt; 0$</span> then we will choose <span class="math-container">$q_A$</span> and <span class="math-container">$q_B$</span> such that (setting up the K-T conditions):</p>
</li>
</ul>
<p><span class="math-container">$$
\begin{align*}
&amp;5 - \lambda = 0\\
&amp;2 q_B - \lambda - \mu = 0\\
&amp;\mu q_B = 0\\
&amp;q_A + q_B = q
\end{align*}
$$</span>
Where <span class="math-container">$\lambda$</span> and <span class="math-container">$\mu$</span> are the Lagrange mulitpliers to the constraints <span class="math-container">$q_A + q_B = q$</span> and <span class="math-container">$q_B \ge 0$</span>. (the restriction <span class="math-container">$q_A &gt; 0$</span> is supposed to hold by assumption). We get that:</p>
<ul>
<li><p>If <span class="math-container">$q_B = 0$</span> then <span class="math-container">$q_A = q$</span> and we have a total cost of <span class="math-container">$50 + 5 q$</span>.</p>
</li>
<li><p>If <span class="math-container">$q_B &gt; 0$</span> then <span class="math-container">$\mu = 0$</span> and we get the solution <span class="math-container">$q_B = \frac{5}{2}$</span> and <span class="math-container">$q_A = q - \frac{5}{2}$</span>. Notice that this can only be a solution for <span class="math-container">$q &gt; \frac{5}{2}$</span> (as <span class="math-container">$q_A &gt; 0$</span>). In this case, total costs are <span class="math-container">$50 + 5(q - \frac{5}{2}) + \frac{25}{4} = 5 q + 50 - \frac{25}{4}$</span></p>
</li>
</ul>
<p>So we end up with 3 candidate solutions:</p>
<ul>
<li><span class="math-container">$q_A = 0, q_B = q$</span> and total costs are <span class="math-container">$q^2$</span>.</li>
<li><span class="math-container">$q_A = q, q_B = 0$</span> and total costs are <span class="math-container">$5q + 50$</span></li>
<li>If <span class="math-container">$q &gt; \frac{5}{2}$</span> we have a potential solution with <span class="math-container">$q_A = q - \frac{5}{2}, q_B = \frac{5}{2}$</span> and total costs are <span class="math-container">$5 q + 50 - \frac{25}{4}$</span>.</li>
</ul>
<p>To see which one is best for which levels of <span class="math-container">$q$</span>, notice that the first solution has a higher cost than the second one if:
<span class="math-container">$$
\begin{align*}
&amp;q^2 - 5q - 50 \ge 0,\\
\leftrightarrow &amp;q \ge 10
\end{align*}
$$</span>
Next, notice that the third solution will always have a lower cost than the second option (in case <span class="math-container">$q &gt; \frac{5}{2}$</span>). and the second option beats the first one only if <span class="math-container">$q \ge 10$</span>. So the second option will never be optimal because <span class="math-container">$q \ge 10$</span> implies <span class="math-container">$q &gt; \frac{5}{2}$</span>.</p>
<p>We are left to decide between the first and third option. The first option will have a higher cost than the third option if:
<span class="math-container">$$
\begin{align*}
&amp;q^2 - 5q - (50 - \frac{25}{4}) \ge 0,\\
\leftrightarrow &amp;q \ge \frac{5}{2}(1 + 2 \sqrt{2}).
\end{align*}
$$</span>
So we have that:</p>
<ul>
<li>if <span class="math-container">$q \le \frac{5}{2}(1 + 2 \sqrt{2})$</span> then <span class="math-container">$q_A = 0$</span> and <span class="math-container">$q_B = q$</span>.</li>
<li>if <span class="math-container">$q &gt; \frac{5}{2}(1 + 2 \sqrt{2})$</span> then <span class="math-container">$q_A = \frac{5}{2}$</span> and <span class="math-container">$q_B = q - \frac{5}{2}$</span>.</li>
</ul>

CC BY-SA 4.0
2022-01-11T07:36:31.197Z
2022-01-11T07:36:31.197Z
tdm
<p>If a firm has the choice of using two production technologies to produce the same output:</p>
<p><strong>A</strong>, featuring a quasi-fixed cost of 50 (i.e. 50 for all q &gt; 0, 0 when q = 0), then a variable cost of 5q</p>
<p><strong>B</strong>, which has no fixed costs but a variable cost of q^2 (so MC = 2q)</p>
<p>Then would the firm's optimal strategy be to continue using <strong>B</strong> until the marginal cost of using <strong>B</strong> equals 50, and only then consider using <strong>A</strong>?</p>

CC BY-SA 4.0
2022-01-11T03:23:05.850Z
On the one hand, I can see why this would work - it's like the firm is considering what technology to use for q = 1, 2, 3, ... and since the cost to get **A** started is so high, you'd use **B** until its marginal cost is high enough to make the firm indifferent between producing the extra unit with **B**, or switching **A** on and incurring the fixed cost. But if the MC of using **A** is very low, and it's just the high quasi-fixed cost deterring the firm from using **A**, it seems like a poor decision to hold off using **A** for that long. (Apologies if this doesn't make sense...)
ABCBAA
CC BY-SA 4.0
2022-01-11T04:01:25.977Z
Does the firm want to minimize cost or marginal cost?
Giskard
CC BY-SA 4.0
2022-01-12T17:34:51.970Z
I realise that the former is true. Would it then be correct to say that marginal costs don't really matter for the firm's cost minimisation problem?
ABCBAA
CC BY-SA 4.0
2022-01-12T19:50:12.237Z
That depends on what you mean by "*don't really matter*". Clearly marginal costs and (total) costs are related.
Giskard
CC BY-SA 4.0
2022-01-13T02:28:05.167Z
In the sense that the essence of the firm's problem is to minimise average/total costs, so if that requires the firm to choose between several production technologies, we would compare the average/total costs of its options at every level of q and choose accordingly. <-- is this correct? If so, marginal costs don't feature in this explanation at all, so I don't see how looking at MC is critical for this specific task.
ABCBAA
CC BY-SA 4.0
2022-01-13T02:32:42.790Z
At every level of $q$ means an infinite number of possibilities. How do you check each without using some optimization trick? These usually rely on the first derivative, marginal cost, as was also the case in tdm's answer.
Giskard
<microeconomics><cost><producer-theory>
Quasi fixed costs - two technologies
ABCBAA
# Quasi fixed costs - two technologies

(Asked by: username_0 on 2022-01-11)

If a firm has the choice of using two production technologies to produce the same output:

**A**, featuring a quasi-fixed cost of 50 (i.e. 50 for all q > 0, 0 when q = 0), then a variable cost of 5q

**B**, which has no fixed costs but a variable cost of q^2 (so MC = 2q)

Then would the firm's optimal strategy be to continue using **B** until the marginal cost of using **B** equals 50, and only then consider using **A**?

**Tags:** microeconomics, cost, producer-theory
> **Comments:**
> **username_0**: On the one hand, I can see why this would work - it's like the firm is considering what technology to use for q = 1, 2, 3, ... and since the cost to get **A** started is so high, you'd use **B** until its marginal cost is high enough to make the firm indifferent between producing the extra unit with **B**, or switching **A** on and incurring the fixed cost. But if the MC of using **A** is very low, and it's just the high quasi-fixed cost deterring the firm from using **A**, it seems like a poor decision to hold off using **A** for that long. (Apologies if this doesn't make sense...)  
> **username_1**: Does the firm want to minimize cost or marginal cost?  
> **username_0**: I realise that the former is true. Would it then be correct to say that marginal costs don't really matter for the firm's cost minimisation problem?  
> **username_1**: That depends on what you mean by "*don't really matter*". Clearly marginal costs and (total) costs are related.  
> **username_0**: In the sense that the essence of the firm's problem is to minimise average/total costs, so if that requires the firm to choose between several production technologies, we would compare the average/total costs of its options at every level of q and choose accordingly. <-- is this correct? If so, marginal costs don't feature in this explanation at all, so I don't see how looking at MC is critical for this specific task.  
> **username_1**: At every level of $q$ means an infinite number of possibilities. How do you check each without using some optimization trick? These usually rely on the first derivative, marginal cost, as was also the case in tdm's answer.  

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## 1 answers

### Answer by username_2 on 2022-01-11. Score: 2 | Accepted answer

We need to solve the following problem:

$$
\min\_{q\_A, q\_B}\left\{50 \times I\_{q\_A > 0} + 5 q\_A + q\_B^2\right\} \text{ s.t. } q\_A + q\_B = q \text{ and } q\_A, q\_B \ge 0
$$
We split up according to whether $q\_A = 0$ or $q\_A > 0$.

* If $q\_A = 0$ then $q\_B = q$ and we have a total cost of $q^2$.
* If $q\_A > 0$ then we will choose $q\_A$ and $q\_B$ such that (setting up the K-T conditions):

$$
\begin{align\*}
&5 - \lambda = 0\\
&2 q\_B - \lambda - \mu = 0\\
&\mu q\_B = 0\\
&q\_A + q\_B = q
\end{align\*}
$$
Where $\lambda$ and $\mu$ are the Lagrange mulitpliers to the constraints $q\_A + q\_B = q$ and $q\_B \ge 0$. (the restriction $q\_A > 0$ is supposed to hold by assumption). We get that:

* If $q\_B = 0$ then $q\_A = q$ and we have a total cost of $50 + 5 q$.
* If $q\_B > 0$ then $\mu = 0$ and we get the solution $q\_B = \frac{5}{2}$ and $q\_A = q - \frac{5}{2}$. Notice that this can only be a solution for $q > \frac{5}{2}$ (as $q\_A > 0$). In this case, total costs are $50 + 5(q - \frac{5}{2}) + \frac{25}{4} = 5 q + 50 - \frac{25}{4}$

So we end up with 3 candidate solutions:

* $q\_A = 0, q\_B = q$ and total costs are $q^2$.
* $q\_A = q, q\_B = 0$ and total costs are $5q + 50$
* If $q > \frac{5}{2}$ we have a potential solution with $q\_A = q - \frac{5}{2}, q\_B = \frac{5}{2}$ and total costs are $5 q + 50 - \frac{25}{4}$.

To see which one is best for which levels of $q$, notice that the first solution has a higher cost than the second one if:
$$
\begin{align\*}
&q^2 - 5q - 50 \ge 0,\\
\leftrightarrow &q \ge 10
\end{align\*}
$$
Next, notice that the third solution will always have a lower cost than the second option (in case $q > \frac{5}{2}$). and the second option beats the first one only if $q \ge 10$. So the second option will never be optimal because $q \ge 10$ implies $q > \frac{5}{2}$.

We are left to decide between the first and third option. The first option will have a higher cost than the third option if:
$$
\begin{align\*}
&q^2 - 5q - (50 - \frac{25}{4}) \ge 0,\\
\leftrightarrow &q \ge \frac{5}{2}(1 + 2 \sqrt{2}).
\end{align\*}
$$
So we have that:

* if $q \le \frac{5}{2}(1 + 2 \sqrt{2})$ then $q\_A = 0$ and $q\_B = q$.
* if $q > \frac{5}{2}(1 + 2 \sqrt{2})$ then $q\_A = \frac{5}{2}$ and $q\_B = q - \frac{5}{2}$.


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