MARGINAL FUNCTIONS IN ECONOMICS: APPLICATION OF DIFFERENTIATION

Economists usually use the following terms:

1. C(x)= Cost of producing x items

2. R(x)=Revenue from the sales of x items

3. P(x)=Profit from sales and production. Of x items

Additionally, economists used the term "marginal" to describe the rate of change.

To put it another way, the term "marginal" means "derivative." Therefore,

C'(x)=Marginal cost

R'(x)=Marginal revenue

P'(x)=Marginal profit.

Example 1

Suppose the cost function is given as $C(x)=0.00002x^3-0.48x^2+300x+150,000$. Determine

1. The marginal cost function

2. The marginal cost of the 7th units

3. The total cost of the 7th unit.

Solution:

1. When economists talk about marginal cost, they mean you should take the derivative of the cost function

$C(x)=0.00002x^3-0.48x^2+300x+150,0000$

Using the power rule of differentiation

$C'(x)=0.00006x^2-0.96x+300$

2. The marginal cost of the 7th unit is computed by substituting 7 for x in the cost function

$C'(7)=0.00006(7)^2-0.96(7)+300$

$C'(7)=0.00294-6.72+300$

$C'(7)=293.28294$

3. $C(x)=0.00002x^3-0.48x^2+300x+150,0000$

$C(7)=0.00002(7)^3-0.48(7)^2+300(7)+150,000$

$C(7)=152,076.48686$

Example 2

If the demand function is

$p(x)=120-3x$.

Determine

1. the marginal revenue function

2. The marginal revenue of the 10th unit

Solution:

1. Recall that total revenue is price times quantity

Hence

$R(x)=(p(x))x$

$R(x)=(120-3x)x$

$R(x)=120x-3x^2$

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

$R'(x)=120-6x$

2. The marginal revenue for the 10th unit is

$R'(x)=120-6(10)$

$R'(x)=60$

Economists also use the term average revenue to refer to the per-unit revenue obtained from selling units of a commodity.

It is calculated as revenue function divided by quantity.

$AR=\frac{R(x)}{x}$

The term average cost is frequently used by economics to refer to the per-unit cost incurred from selling units of a commodity.

It is calculated as cost function divided by the quantity

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

Example 3

Given that cost function is

$C(x)=2x^2+6x+13$

Determine

1. The average cost

2. The marginal cost

3. The fixed cost

4. The variable cost of the 5th unit

Solution:

1. $AC=\frac{C(x)}{x}$

$AC=\frac{2x^2+6x+13}{x}$

$AC=\frac{2x^2}{x}+\frac{6x}{x}+\frac{13}{x}$

$AC=2x+6+\frac{13}{x}$

2. The marginal cost expression is the first derivative of the cost function

$C'(x)=4x+6$

3. The fixed cost is the total cost when the production level is zero

When x=0

$C(x)=2(0)^2+6(0)+13$

$C(x)=13$

4. The variable cost is the part of the cost function that has a variable.

Hence

$VC=2x^2+6x$

At the fifth unit,

$VC=2(5)^2+6(5)$

$VC=50+30=80$

As earlier noted, economists use P(x)to denote profit.

Profit is defined as the difference revenue between costs. Therefore,

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

Average profit is the per-unit profit obtained from the sales and production of goods and services.

It is computed as the profit function divided by quantity.

$$AP=\frac{P(x)}{x}$$

Furthermore, Economists also calculate marginal profit.

Marginal profit is the additional profit obtained from selling a unit of a good.

It is the first derivative of the profit function.

It should be noted that a firm maximizes profit at the production level where marginal profit is zero.

Example 4

Kenny knows that his cost function is $C(x)=5000+600x$ and the Revenue function is $R(x)=-\frac{x^2}{2}+1000x$. Determine

1. The profit function

2. The production level that maximizes profit

3. Maximum profit that the firm can earn

Solution:

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

$P(x)=-\frac{x^2}{2}+1000x-(5000+600x)$

$P(x)=-\frac{x^2}{2}+1000x-5000-600x$

$P(x)=-\frac{x^2}{2}+400x-5000$

2. A firm maximizes profit at the production level where marginal profit is zero.

$P(x)=-\frac{x^2}{2}+400x-5000$

Marginal profit is the first derivative of the profit function, hence,

$P'(x)=\frac{-2x}{2}+400$

$P'(x)=-x+400$

Because a firm maximizes profit where marginal profit is zero

$-x+400=0$

$-x=-400$

$x=400$

Therefore, the production level that maximizes profit is the 400th unit

3. Since the 400th unit maximizes profit, it follows that the maximum profit possible will be the profit at the 400th unit.

$P(x)=-\frac{x^2}{2}+400x-5000$

$P(400)=-\frac{(400)^2}{2}+400(400)-5000$

$P(400)=-\frac{160000}{2}+160000-5000$

$P(400)=75,000$

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