<p>The total cost was 291 m.u. for 1 unit given $MC(q) \equiv \frac{dC(q)}{q} = 3 q^2 - 40 q + 220$.</p>

<p>First, the area under the Marginal Cost gives the Total Variable Cost, $TVC(q)$. To find this area, we will integrate the Marginal Cost:</p>

<p>$$
\begin{align}
TVC(q) &amp;= \int_0^q\frac{dTVC(q)}{dq} dq = \int_0^q (3q^2 - 40 q + 220) dq\\
&amp;= \left. \left(q^3 -20q^2 +220q  \right) \right|_q - \left. \left(q^3 -20q^2 +220q\right)  \right|_0\\
&amp;= {q^3 -20q^2 +220q}
\end{align}
$$
Now, you have to use the additional information of 291 m. u. to find the Fixed Cost. The Total Cost equals the Variable Cost plus the Fixed Cost (which does not depend on q):
$$\begin{align}
TC(1) = TVC(1) + FC &amp;=  291\\ 
1^3 -20 \cdot 1^2 + 220 \cdot 1 + FC &amp;= 291\\
FC &amp;= 90
\end{align}
$$</p>

<p>Hence, for the 8 units:
$$
\begin{align}
TC(8) = TVC(8) + FC &amp;= 8^3-20 \cdot 8^2 + 220 \cdot 8 + 90\\
&amp;= 1082
\end{align}
$$</p>

CC BY-SA 3.0
2015-06-03T19:43:40.623Z
2015-06-03T23:08:33.187Z
2015-06-03T23:08:33.187Z
Konstantinos
<p>$MC = (TC)'(q) = 3q^2 - 40q+220, \quad  where \quad  1 \leq q \leq 20 $</p>

<p>The task is to deduce a function of $TC$, if for the production of the first item we need 291 monetary units, and to write out the cost value if $q$ = 8 units.</p>

<p>I have found that $TC = q^3-20q^2+220q$, but i can't understand how to use that 291 monetary units for finding the cost value</p>

CC BY-SA 3.0
2015-06-03T17:30:00.067Z
We tend to close homework question with no personal work over here. Please edit after trying by yourself.
VicAche
CC BY-SA 3.0
2015-06-03T18:23:02.600Z
This will help you:
http://en.wikipedia.org/wiki/Cost_curve#Relationship_between_different_curves
(Especially the first line.)
Giskard
<cost>
Deducing the Total Cost function
uiuiui000
# Deducing the Total Cost function

(Asked by: username_0 on 2015-06-03)

$MC = (TC)'(q) = 3q^2 - 40q+220, \quad where \quad 1 \leq q \leq 20 $

The task is to deduce a function of $TC$, if for the production of the first item we need 291 monetary units, and to write out the cost value if $q$ = 8 units.

I have found that $TC = q^3-20q^2+220q$, but i can't understand how to use that 291 monetary units for finding the cost value

**Tags:** cost

> **Comments:**
> **username_1**: We tend to close homework question with no personal work over here. Please edit after trying by yourself.  
> **username_2**: This will help you: http://en.wikipedia.org/wiki/Cost_curve#Relationship_between_different_curves (Especially the first line.)  

---

## 1 answers

### Answer by username_3 on 2015-06-03. Score: 3

The total cost was 291 m.u. for 1 unit given $MC(q) \equiv \frac{dC(q)}{q} = 3 q^2 - 40 q + 220$.

First, the area under the Marginal Cost gives the Total Variable Cost, $TVC(q)$. To find this area, we will integrate the Marginal Cost:

$$
\begin{align}
TVC(q) &= \int\_0^q\frac{dTVC(q)}{dq} dq = \int\_0^q (3q^2 - 40 q + 220) dq\\
&= \left. \left(q^3 -20q^2 +220q \right) \right|\_q - \left. \left(q^3 -20q^2 +220q\right) \right|\_0\\
&= {q^3 -20q^2 +220q}
\end{align}
$$
Now, you have to use the additional information of 291 m. u. to find the Fixed Cost. The Total Cost equals the Variable Cost plus the Fixed Cost (which does not depend on q):
$$\begin{align}
TC(1) = TVC(1) + FC &= 291\\
1^3 -20 \cdot 1^2 + 220 \cdot 1 + FC &= 291\\
FC &= 90
\end{align}
$$

Hence, for the 8 units:
$$
\begin{align}
TC(8) = TVC(8) + FC &= 8^3-20 \cdot 8^2 + 220 \cdot 8 + 90\\
&= 1082
\end{align}
$$


---

