Minimization of iron costs: financial mathematics for an high school Meanwhile I apologize to everyone. I have never treated financial mathematics. A colleague of mine asked me for help but I can't give an answer. I have two exercise. This is the first.

A company operates in the iron sector, and sustains the following fixed costs equal to $ 648 \, €$ for month. A variable cost of $4$ € euros per quintal produced. An additional cost determined by the use of a particular ironworking machinery, equal to half of the square of the quintals produced.


*

*What are the function of the total monthly cost and the function of the average cost?

*What is the production at which you have the lowest average cost and what is the total monthly cost at this quantity?

The results are:
$$a) \quad C=0.5q^2+4q+648, \quad C_m=0.5q+\frac{648}q+4$$
$$b)\quad  q=36\, \frac{\text{quintals}}{\text{month}};\quad  C(36)=€\, 1440$$

My observation: with the first informations of the problem I have a parabola. Why? For the rest of the questions I think that there are some derivative to be done. Never studied financial mathematics. Thank you all for your understanding.
 A: Before to understand the exercise I had to study a bit on the internet. Regarding the average cost $C_m(q)$ one must keep in mind that the average cost is equal to the cost per unit of production; while the marginal cost $C_{\text{mar}}$ represents the change in cost due to a unit increase in production which we will denote by the independent variable $q$. We therefore have:
\begin{equation}
    C_m(q)=\frac{C(q)}{q}
\end{equation}
and
\begin{equation}
    C_{\text{mar}}(q)=\frac{dC(q)}{dq}
\end{equation}
So in Economics we use linear, quadratic, cubic, exponential etc. functions for the cost function. The curves of the average cost and marginal cost intersect at the stationary points of the average cost (i.e. at the points where the average cost has zero derivative).
Given the nature of the problem with the indications provided of the exercise we have three parameters. Excluding the circumference which is not a function unless there are appropriate domain restrictions, the simplest case is represented by a parabola.
$$C(q)=0.5q^2+4q+648$$
with average cost equal to
$$C_m(q)=\frac{0.5q^2+4q+648}{q}=0.5q+\frac{648}{q}+4$$
Now for the marginal cost it is necessary to use the derivative, equal to
$$C_{\text{mar}}=\frac{dC(q)}{dq}=q+4$$
Intersecting the two functions unit or average cost $C_m(q)=(0.5q^{2}+4x+648)/q$, branch of nonequilateral hyperbola, and marginal cost we have the intersection point $(36, 40)$, so $q=36\, \text{quintals}/\text{month}$ and $C(36)=1440$ euros.
