- Problem
- Detailed Solution
- Summary Solution

### Determining parameters of a cost function

Find a third degree polynomial cost function

\(C=C(x)= a{x^3} + b{x^2} + cx + d\)

that has the following properties:

- fixed cost is 20 CHF
- minimal marginal cost is 0.3 CHF/qu
- marginal cost is minimal when the quantity produced is 40 qu
- when 40 qu are being produced, average cost is 2 CHF/qu.

### Problem structure and preliminary remarks:

The paramaters a, b, c, and d have to be determined so that C has the properties described above. This will lead to a system consisting of 4 equations for the four unknown paramater values.

To solve this problem, you need the following competences:

- Knowledge on polynomial functions (see block I learning sequence 2 of our script)
- Knowledge on the terms of cost theory used above
- Knowledge on linear equation systems and how to solve them using matrix theory (see block IV of our script)

### Formal approach:

A cost function is the sum of fixed and variable cost:

\(C = {C_f} + {C_v}\)

In our case, we have \(\;{C_f}=d=20\).

Thus, we already found one of the three parameter values we have been looking for (using one almost trivial equation), and we can restate our cost function equation as follows:

\(C(x) = a{x^3} + b{x^2} + cx + 20\)

Since average cost is total cost per quantity unit, all we have to do to obtain that function is divide C by x:

\(c(x) = \frac{{C(x)}}{x} = \frac{{a{x^3} + b{x^2} + cx + d}}{x} = a{x^2} + bx + c + \frac{d}{x}\)

(Remark: average functions usually get the name of the original function, but with small letters).

We can now express the last condition as an equation: «Average
cost is 2 CHF/qu when the quantity produced is 40 qu» means
nothing else but \(c(40) = 2\), and we get the equation

\(c(40) = {40^2} \cdot a + 40 \cdot b + c + \frac{{20}}{{40}} = 2\quad \Rightarrow \quad 1600a + 40b + c + 0.5 = 2\)

This is our **first equation**.

Marginal cost is the 1^{st} order derivative of the
(total) cost function C:

\(C'(x) = 3a{x^2} + 2bx + c\)

Conditions 2 and 3 state that marginal cost is 0.3 when the quantity is 40, which means that

\(C'(40) = 3 \cdot {40^2} \cdot a + 2 \cdot 40 \cdot b + c = 0.3\quad \Rightarrow \quad 4800a + 80b + c = 0.3\)

This is our **second equation**.

Condition 3 implies yet another information: since x = 40 is the
place where marginal cost is minimal, that function must itself
have a critical value of 40, and we conclude that the 1^{st}
order derivative of C' – which is nothing else but C'' – must be 0
there.

With \(\;C''(x) = 6ax + 2b\,\), we have

\(C''(40) = 6 \cdot 40 \cdot a + 2 \cdot b = 0 \quad \Rightarrow \quad 240a + 2b = 0\).

This is our **third equation**, and since there
remain three parameter values to be determined, these equations
may allow us to to so.

We restate our three equations as a system (please keep in mind that, since we already know that d = 20, we have three instead of four unknown parameter values, and three indstead of four equations):

\(4800a + 80b + c = 0.3\)

\(240a + 2b = 0\)

\(1600a + 40b + c + 0.5 = 2\)

This obviously is a linear equation system (LES), and it can be stated in matrix form (see block IV learning sequence 2 of our script):

\(\left( {\begin{array}{*{20}{c}}{4800}&{80}&1\\{240}&2&0\\{1600}&{40}&1\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}a\\b\\c\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{0.3}\\0\\{1.5}\end{array}} \right)\)

The solution can be obtained like this:

\(\left( {\begin{array}{*{20}{c}}a\\b\\c\end{array}} \right) = {\left( {\begin{array}{*{20}{c}}{4800}&{80}&1\\{240}&2&0\\{1600}&{40}&1\end{array}} \right)^{ - 1}} \cdot \left( {\begin{array}{*{20}{c}}{0.3}\\0\\{1.5}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{0.00075}\\{ - 0.09}\\{3.9}\end{array}} \right)\)

Since we now know all the parameter values, we have determined our cost function, which we state as follows:

\(C(x) = 0.00075{x^3} - 0.09{x^2} + 3.9x + 20\)

See diagram.

Concluding remark:

Please check whether this function really
has all the required properties. The reason for this is that we
made use of necessary, not of sufficient conditions.

Fixed cost is 20, so we have

\(C(x) = a{x^3} + b{x^2} + cx + 20\)

Derivatives:

\(C'(x) = 3a{x^2} + 2bx + c\)

\(C''(x) = 6ax + 2b\)

Average cost:

\(c(x) = \frac{{C(x)}}{x} = \frac{{a{x^3} + b{x^2} + cx + d}}{x} = a{x^2} + bx + c + \frac{d}{x}\)

The conditions now read:

\(C'(40) = 0.3 \quad \Rightarrow \quad 4800a + 80b + c = 0.3\)

\(C''(40) = 0 \quad \Rightarrow \quad 240a + 2b = 0\)

\(c(40) = 2 \quad \Rightarrow \quad 1600a + 40b + c + 0.5 = 2\)

Since these are linear equations, we can put them together as a LES (don't forget to standardize the last one!):

\(\left( {\begin{array}{*{20}{c}}{4800}&{80}&1\\{240}&2&0\\{1600}&{40}&1\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}a\\b\\c\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{0.3}\\0\\{1.5}\end{array}} \right)\)

\(\left( {\begin{array}{*{20}{c}}a\\b\\c\end{array}} \right) = {\left( {\begin{array}{*{20}{c}}{4800}&{80}&1\\{240}&2&0\\{1600}&{40}&1\end{array}} \right)^{ - 1}} \cdot \left( {\begin{array}{*{20}{c}}{0.3}\\0\\{1.5}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{0.00075}\\{ - 0.09}\\{3.9}\end{array}} \right)\)

Thus, we obtain the equation of the cost function:

\(C(x) = 0.00075{x^3} - 0.09{x^2} + 3.9x + 20\)