COST FUNCTION CALCULATIONS: DERIVING MARGINAL, AVERAGE COST, AVERAGE FIXED COST


Most times, economists do not give the exact cost of producing goods. Instead, they represent it as a function. This function is called the cost function.

At the risk of oversimplifying, a cost function is a functional relationship between all costs of inputs used in the production of a good.

Deriving cost definitions(such as fixed cost, marginal cost, variable cost) from cost functions is one problem faced by many students.

For this reason, we are going to learn how to derive various cost definitions from cost functions using these three examples.

Example 1
The total cost(€) of producing books is given as $C(Q)=0.05(q)^3-0.02(q)^2-30q+3000$. Find the total cost of producing 6 units

1. Find the total cost of producing 10 units
2. Find the average cost of producing that same unit

Solution:
$C(q)=0.05(q)^3-0.02(q)^2-30q+3000$

By substitution
$C(10)=0.05(10)^3-0.02(10)^2-30(10)+3000$
          $=50-2-300+3000$
          $=2748$

Therefore the total cost of producing the $10$ units of books is $€2748$

The average cost can easily be computed by dividing total cost by quantity
$ATC=\frac{2748}{10}$
$ATC=274.8$

Example 2
A book-producing firm cost function is given as $C(Q)=300+4q$. Find
1. The total cost of producing the fifth book
2. The fixed cost of producing the fifth book
3. The variable cost of producing that same unit

Solution:
$C(Q)=300+4q$

By substitution
$C(5)=300+4(5)$
          $=320$

The total cost of producing the fourth unit is €320.

$FC(Q)=300+4(0)$
$FC(Q)=300$

Therefore, the fixed cost is $300$

The variable cost can easily be computed by subtracting the total cost from the fixed cost. Accordingly,

$VC=320-300$
$VC=20$

Another way of finding the variable cost is to use the variables in the cost function ($C(Q)=300+4q$). 

In the above cost function, $300$ is a constant while $4q$ is a variable. Accordingly,
$VC(Q)=4Q$

By substitution
$VC(Q)=4(5)$
$VC(Q)=20$

Therefore, the variable cost of producing the 5th units is $20$.

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Example 3
Given the cost function $C(Q)=100+10Q+Q^2$. 
1. Find the quantity that has the lowest per cost
2.  Find the fixed cost of producing that quantity
3. Also find the total cost of producing that unit

Solution:
The lowest per-unit cost means the lowest average cost. 

To find the quantity that has the lowest average cost, we equate Marginal cost and average cost. Remember average cost will be lowest when marginal cost is equal to the average cost

$MC=ATC$


$C(Q)=100+10Q+Q^2$

Taking the derivatives of the cost function, this translates to:
$C'(Q)=10+2Q$
$MC(Q)=10+2Q$.

The average cost is easily computed by divided total cost by quantity. Accordingly,
$AC(Q)=\frac{100+10Q+Q^2}{Q}$

Now, let substitute the function of $MC$ and $AC$ in $MC=ATC$

$10+2Q=\frac{100+10Q+Q^2}{Q}$

Cross multiply
$Q(10+2Q)=100+10Q+Q^2$
$10Q+2Q^2=100+10Q+Q^2$
$2Q^2-Q^2+10Q-10Q=100$
$Q^2=100$

Adding square root to both sides
$\sqrt{Q^2}=\sqrt{100}$
$Q=10$

Therefore, the 10th quantity has the lowest per-unit cost.

Fixed cost is constant. It does not vary with output. And at zero output, all cost are fixed. Accordingly, 

$C(Q)=100+10Q+Q^2$
$C(0)=100+10(0)+(0)^2$
         $=100$.
Since fixed cost does not vary. We can conclude that the fixed cost of producing the 10th quantity is $100$

To derive total cost, we simply insert $10$ for $Q$ in the cost function

$C(Q)=100+10Q+Q^2$
$C(10)=100+10(10)+(10)^2$
$C(10)=100+100+100$
            $=300$

Hence, the total cost of producing the $10$ unit is 300.

Ok, that will be all for now. In this next post, we will be using the idea of cost functions to calculate profit-maximization price and output.

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