# BUSINESS FUNCTIONS (COST, REVENUE, AVERAGE COST, PRICE, DEMAND)(CALCULUS)

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Various functions are used in economics and business. In today's post, we will be focusing on just a few of them.

## Cost function

The cost function represents the total cost function of producing a particular number of units of an item.

The total cost of producing any units of an item is simply the sum of fixed cost and variable cost.

Mathematically, a cost function is expressed as:

$C(x)=Vc(x)+Fc(x)$

From the above, Vc(x) represents the variable cost function, which is the variable cost of producing x number of items.

Fc(x) represents the fixed cost, which is usually constant.

For example, if we have a cost function of $C(x)=5x^2+6x+10$, then the fixed cost is 10 because it is a constant, and the variable cost will be $Vc(x)= 5x^2+6x$.

## Average cost function

The average cost function represents the per unit cost of producing a particular unit of a commodity. The average cost is the total cost function divided by quantity.

Thus, the average cost is expressed as

$c(x)=\frac{C(x)}{x}$

Please note that the average cost function is denoted by the small letter c while the cost function is denoted by capital C

## Marginal cost function

The marginal cost function represent cost incurred from producing an additional unit of a product.

The marginal cost is simply the first derivative of the cost function. So, the marginal cost function is denoted by:

$C'(x)$

Where $C'(x)$ is the first derivative of the cost function.

## Minimization Rule

To minimize the per unit cost (average cost) of production, a firm must produce where the average cost is equal to the marginal cost.

So, to minimize average cost, a firm must produce at the output where the average cost function equals the marginal cost function. That is,

$c(x)=C'(x)$

Another way of minimizing cost is to take the derivative of the average cost function and then equate it to zero.

$c'(x)=0$

To put what we have discussed so far to practice, let's take an example

### Example 1

Supposed the cost function of a firm is given as $C(x)=500000+1500x+x^2$. Find

i. The total cost of producing 10 units of output

ii. The average cost of producing 10 units of output

iii. The marginal cost of producing 10 units of output

iv. The production level that minimizes average cost

v. Find the minimum average cost

Solution:

i. The total cost function is given as $C(x)=500000+1500x+x^2$

Hence, the cost function of producing 10 units is

$C(10)=500000+1500(10)+(10)^2$

$C(10)=515,100$

So, the total cost of producing 10 units is N515,100.

ii. Recalled that average cost function is: $c(x)=\frac{C(x)}{x}$,

For the 10 units,

$c(x)=\frac{C(10)}{10}$

We already know that C(10) is 501,600. Accordingly,

$c(x)=\frac{515,100}{10}=51,510$

So, the cost per unit is N51,510

iii.  Marginal cost is the first derivative of the cost function.

Taking the derivative

$C'(x)=1500+2x$

When x=10

$C'(10)=1500+2(10)$

$C'(10)=1500+20$

$C'(10)=1520$

iv. I earlier noted that the production level that minimizes average cost is the quantity where average cost equals marginal cost.

If total cost is $C(x)=500000+1500x+x^2$, then

$c(x)=\frac{500000+1500x+x^2}{x}$ and $C'(x)=1500+2x$

Equating average cost function to marginal cost,

$\frac{500000+1500x+x^2}{x}=1500+2x$

By cross multiplication

$500000+1500x+x^2=1500x+2x^2$

$500000=2x^2+x^2$

$500000=x^2$

$x=707$

So, the production level that minimizes per unit is roughly 707 units.

v.  The minimum average cost will be the average cost of the 707 units when the average cost is minimum

$c(x)=\frac{500000+1500x+x^2}{x}$

$c(707)=\frac{500000+1500(700)+(700)^2}{700}$

$c(707)=2914$

The minimum average cost is approximate N2914

## Price function

The price function indicates the price that a firm can charge per unit of a good. The price function is also known as the inverse demand function.

According to the law of demand, the price of a commodity rises as the quantity and vice versa.

So, it is expected that the quantity demanded(x) will fall as the price function rises and rise when the price function falls.

The price function is denoted using lowercase p, like this: $p(x)$.

## Revenue function

This represents the total revenue earned by the firm for selling a particular unit of a commodity.

Since total revenue is price times quantity, Then revenue function is represented as follows:

$R(x)=x \times p(x)$

## Marginal revenue function

The marginal revenue is derived by taking the first derivative of the revenue function.

The marginal revenue shows the additional revenue that a firm can realize by selling an additional unit of a product.

It is denoted as $R'(x)$.

## Revenue maximization rule

According to the revenue maximization rule, a firm maximizes revenue when marginal revenue is equal to zero.

So, to maximize revenue, we must equate marginal revenue to zero

### Example 2

Given that the price function of firm is$p(x)=3x-90$ , Determine

i. the revenue function

ii. the marginal revenue function

iii. the production level that maximizes revenue

Solution:

i. $p(x)=3x-90$

Recall that $R(x)=x \times p(x)$

$R(x)=x (3x-90)$

$R(x)=3x^2-90x$

ii. The marginal revenue function is the first derivative of the revenue.

$R(x)=3x^2-90x$

Taking the derivative

$R'(x)=6x-90$

iii.  The production level that maximizes revenue means we should find the quantity that maximizes revenue. So, we equate the marginal revenue function to zero.

$6x-90=0$

$6x=90$

$x=15$

## Profit Function

Profit, as you might be aware, is the difference between revenue and cost.

Therefore, the profit function, denoted by P(x), is obtained by subtracting the cost function from the revenue function.

That is,

$P(x)=R(x)-C(x)$

Note that we denote the profit function using capital P, as opposed to the small letter used in denoting abe obtained price function.

## Marginal profit function

This shows the additional profit a firm earned from selling additional units of goods.

The marginal profit can be obtained using two methods.

It can be obtained by taking the derivative of the profit function or by subtracting marginal cost from marginal revenue.

Marginal Profit is denoted using $P(x)$

$P'(x)=R'(x)-C'(x)$

## Profit maximization rule

The profit maximization rule states that a firm maximizes profit where marginal profit is zero or where marginal revenue equals marginal cost.

If we are to express it symbolically, profit is maximized when:

$P'(x)=0$ or $R'(x)=C'(x)$

### Example 3

Given that the cost function of company XYZ is $C(x)=1000+5x+0.01x^2$ and the price function is $p(x)=20$. Determine

i. the revenue function

ii. the marginal revenue function

iii. the marginal cost function

iv. The production level that maximizes profit

v. Determine the maximum profit that the company can earn

Solution:

i. Recalled that $R(x)=x \times p(x)$

Hence,

$R(x)=x(20)$

$R(x)=20x$

ii. the marginal revenue function is derived by taking the derivative of the revenue function.

$R'(x)=20$

iii. The marginal cost function is obtained by taking the derivative of the cost function

$C(x)=1000+5x+0.01x^2$

$C'(x)=5+0.02x$

iv. I told you earlier that a firm maximizes revenue when marginal revenue is equal to marginal cost.

So, equating MR to MC,

$20=5+0.02x$

$20-5=0.02x$

$15=0.02x$

$\frac{15}{0.02}=x$

$x=750$

v.  Profit is the difference between revenue and cost. So,

$P(x)=R(x)-C(x)$

Inserting the revenue and cost functions

$P(x)=20x-(1000+5x+0.01x^2)$

$P(x)=20x-1000-5x-0.01x^2$

Since we know the firm earns the maximum profit where the quantity is 750, it follows that the maximum profit the firm can earn is the profit of the 750th units.

$P(750)=20(750)-1000-5(750)-0.01(750)^2$

$P(750)=4625$

That will be all for now. I hope you can calculate the major business functions used by economics.

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