Applications of Derivatives in Economics

Neetesh Kumar | February 08, 2025 $\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space$ Share this Page on:

• 1. Introduction to Application of Derivatives in Economics
• 2. Formula Used
• 3. Solved Examples
• 4. Conclusion

Derivatives play a crucial role in economics by helping us understand how functions change over time or under varying conditions.

In particular, derivatives are used to model various economic phenomena such as:

• Marginal Cost and Revenue: The rate of change of cost and revenue with respect to quantity.
• Elasticity of Demand: The responsiveness of the quantity demanded to price changes.
• Optimization: Maximizing profit or minimizing cost by analyzing the behavior of economic functions.

1. Marginal Cost and Revenue:

Marginal cost (MC) and marginal revenue (MR) are the derivatives of the total cost and total revenue functions, respectively:

• Marginal Cost (MC): $MC = \dfrac{dC}{dq}$
• Marginal Revenue (MR): $MR = \dfrac{dR}{dq}$

Where:

• $C$ is the total cost function.
• $R$ is the total revenue function.
• $q$ is the quantity of goods produced and sold.

2. Elasticity of Demand:

The Price Elasticity of Demand (PED) measures the responsiveness of the quantity demanded to a change in price:

$E_d = \dfrac{dQ}{dP} \times \dfrac{P}{Q}$

Where:

• $Q$ is the quantity demanded.
• $P$ is the price.
• $\dfrac{dQ}{dP}$ is the derivative of the demand function.

3. Profit Maximization:

Profit is the difference between total revenue and total cost:

$\text{Profit} = R(q) - C(q)$

To maximize profit, we take the derivative of the profit function and set it equal to zero:

$\dfrac{d\text{(Profit)}}{dq} = \dfrac{dR}{dq} - \dfrac{dC}{dq} = 0$

This gives the quantity $q^*$ that maximizes profit.

Example 1: Marginal Revenue and Cost

The total cost function is $C(q) = 3q^3 + 5q^2 + 2q + 10$, and the total revenue function is $R(q) = 20q^2 + 50q$. Find the marginal cost and marginal revenue at $q = 2$.

Solution:

To find the marginal cost and marginal revenue, we differentiate the total cost and revenue functions:

• Marginal Cost: $MC = \dfrac{dC}{dq} = 9q^2 + 10q + 2$
• Marginal Revenue: $MR = \dfrac{dR}{dq} = 40q + 50$

Substitute $q = 2$ into both expressions:

• $MC(2) = 9(2)^2 + 10(2) + 2 = 36 + 20 + 2 = 58$
• $MR(2) = 40(2) + 50 = 80 + 50 = 130$

Thus, the marginal cost at $q = 2$ is 58, and the marginal revenue is 130.

Example 2: Elasticity of Demand

The demand function is given by $Q = 50 - 2P$, where $Q$ is the quantity demanded and $P$ is the price. Find the price elasticity of demand at $P = 10$.

Solution:

First, find the derivative of the demand function:

$\dfrac{dQ}{dP} = -2$

Now, substitute $P = 10$ into the demand function to find $Q$:

$Q = 50 - 2(10) = 30$

Now calculate the price elasticity of demand:

$E_d = \dfrac{dQ}{dP} \times \dfrac{P}{Q} = -2 \times \dfrac{10}{30} = -\dfrac{2}{3}$

Thus, the price elasticity of demand at $P = 10$ is $-\dfrac{2}{3}$.

Example 3: Profit Maximization

The total cost function is $C(q) = 2q^2 + 10q + 100$ and the total revenue function is $R(q) = 20q - q^2$. Find the quantity $q^*$ that maximizes profit.

Solution:

Profit is given by:

$\text{Profit} = R(q) - C(q)$

So, the profit function is:

$\text{Profit}(q) = (20q - q^2) - (2q^2 + 10q + 100)$

Simplify:

$\text{Profit}(q) = -3q^2 + 10q - 100$

Now, differentiate the profit function with respect to $q$:

$\dfrac{d\text{(Profit)}}{dq} = -6q + 10$

Set the derivative equal to zero to find the critical points:

$-6q + 10 = 0$

$q = \dfrac{10}{6} = \dfrac{5}{3}$

Thus, the quantity $q^*$ that maximizes profit is $\dfrac{5}{3}$.

Example 4: Marginal Profit Calculation

The cost function is $C(q) = 4q^3 - 5q^2 + 10q$, and the revenue function is $R(q) = 8q^2 + 20q$. Find the marginal profit at $q = 3$.

Solution:

Profit is the difference between revenue and cost:

$\text{Profit}(q) = R(q) - C(q)$

$\text{Profit}(q) = (8q^2 + 20q) - (4q^3 - 5q^2 + 10q)$

Simplify:

$\text{Profit}(q) = -4q^3 + 13q^2 + 10q$

Now, differentiate the profit function with respect to $q$:

$\dfrac{d\text{(Profit)}}{dq} = -12q^2 + 26q + 10$

Substitute $q = 3$:

$\dfrac{d\text{(Profit)}}{dq} = -12(3)^2 + 26(3) + 10 = -108 + 78 + 10 = -20$

Thus, the marginal profit at $q = 3$ is $-20$.