<p>It is the first one, <span class="math-container">$TC(0) = FC$</span>. This is the definition. </p>

<p>Also consider that it is not clear what is "transformed by <span class="math-container">$q$</span> in some way". In case of
<span class="math-container">$$
\frac{5q}{q+1} + \frac{5}{q+1}
$$</span>
are the two fractions transformed by <span class="math-container">$q$</span>, or should I just sum them up to 5?</p>

<p>With your function, one can rearrange it to
<span class="math-container">$$
TC(q) = \frac{5}{q+1} + 5 + 5q + q^2 = -\frac{5q}{q+1} + 10 + 5q + q^2.
$$</span></p>

CC BY-SA 4.0
2020-03-18T13:26:52.647Z
The rearrangement is very clear. Thank you!
pc724
CC BY-SA 4.0
2020-03-18T05:38:26.787Z
2020-03-18T13:40:52.127Z
2020-03-18T13:40:52.127Z
Giskard
<p>Suppose that a firm has a total cost function given by: 
<span class="math-container">$TC(q) = \frac{5}{q+1} + 5 + 5q + q^2$</span>. 
What is the fixed cost?</p>

<p>I seem to be able to come up with two "answers", which cannot be correct. My hunch is that it is because I do not have a precise understanding of what a "fixed cost" is, as silly as that may sound. </p>

<p><strong>Option 1</strong>
Fixed cost is the cost of production when <span class="math-container">$q = 0$</span>. Evaluating <span class="math-container">$TC(0) = 10$</span>. Thus, fixed costs are $10. </p>

<p><strong>Option 2</strong>
Intuitively, fixed costs should be any term in the in the cost function that are not transformed by "q" in some way, since fixed costs should remain "fixed" and independent across any level of production. This would imply that fixed costs are $5. </p>

<p>I think clearly there is something I am missing here since there should be one answer. Can someone help me understand what the correct answer is and where exactly where I am going wrong with my reasoning for the incorrect option? Thank you! </p>

CC BY-SA 4.0
2020-03-18T15:11:23.037Z
If you have an intergrable marginal cost function $MC(q)$ then you can say $TC(q)=FC+\int\limits_{x=0}^q MC(x)\, dx$ and it becomes clear that $FC=TC(0)$.  In this case $MC(q) = 5+2q-\frac{5}{(q+1)^2}$ and $FC=5$
Henry
CC BY-SA 4.0
2020-03-19T16:11:05.727Z
@Henry, You mean $FC = 10$, correct?
pc724
<microeconomics><cost><producer-theory><cost-functions>
Fixed cost of a firm
pc724
# Fixed cost of a firm

(Asked by: username_0 on 2020-03-18)

Suppose that a firm has a total cost function given by:
$TC(q) = \frac{5}{q+1} + 5 + 5q + q^2$.
What is the fixed cost?

I seem to be able to come up with two "answers", which cannot be correct. My hunch is that it is because I do not have a precise understanding of what a "fixed cost" is, as silly as that may sound.

**Option 1**
Fixed cost is the cost of production when $q = 0$. Evaluating $TC(0) = 10$. Thus, fixed costs are $10.

**Option 2**
Intuitively, fixed costs should be any term in the in the cost function that are not transformed by "q" in some way, since fixed costs should remain "fixed" and independent across any level of production. This would imply that fixed costs are $5.

I think clearly there is something I am missing here since there should be one answer. Can someone help me understand what the correct answer is and where exactly where I am going wrong with my reasoning for the incorrect option? Thank you!

**Tags:** microeconomics, cost, producer-theory, cost-functions

> **Comments:**
> **username_1**: If you have an intergrable marginal cost function $MC(q)$ then you can say $TC(q)=FC+\int\limits_{x=0}^q MC(x)\, dx$ and it becomes clear that $FC=TC(0)$.  In this case $MC(q) = 5+2q-\frac{5}{(q+1)^2}$ and $FC=5$  
> **username_0**: @username_1, You mean $FC = 10$, correct?  

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## 1 answers

### Answer by username_2 on 2020-03-18. Score: 10 | Accepted answer

It is the first one, $TC(0) = FC$. This is the definition.

Also consider that it is not clear what is "transformed by $q$ in some way". In case of
$$
\frac{5q}{q+1} + \frac{5}{q+1}
$$
are the two fractions transformed by $q$, or should I just sum them up to 5?

With your function, one can rearrange it to
$$
TC(q) = \frac{5}{q+1} + 5 + 5q + q^2 = -\frac{5q}{q+1} + 10 + 5q + q^2.
$$

> **Comments:**
> **username_0**: The rearrangement is very clear. Thank you!  

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