Home Class 11 Applied Maths Chapter 7: Calculus Exercise 7.5: Economic and Practical Applications of Derivatives
Chapter 7 • Unit III

Exercise 7.5: Economic and Practical Applications of Derivatives

Comprehensive guide and practice questions for CBSE Class 11 Applied Mathematics curriculum (2026-27).

7.5 Economic and Practical Applications of Derivatives

In corporate strategy, microeconomics, and managerial accounting, derivatives are not theoretical abstractions—they are the direct mathematical expression of marginality. Whenever an executive asks "What is the cost of manufacturing one more unit?" or "How much revenue is gained by lowering price by 1%?", the answer is computed through differential calculus.

Master Formulary of Economic Derivatives

1. Cost Function Metrics

  • Total Cost: \(C(x) = F + V(x)\), where \(F\) is Fixed Overhead and \(V(x)\) is Variable Cost of \(x\) units.
  • Average Cost (AC): Cost per unit produced: \[ AC(x) = \frac{C(x)}{x} \]
  • Marginal Cost (MC): Instantaneous rate of change of total cost with respect to output: \[ MC(x) = \frac{dC}{dx} = C'(x) \]
  • Average Cost Minimization Criterion: Average cost is minimized at the output level where Marginal Cost equals Average Cost: \[ \frac{d}{dx}[AC(x)] = 0 \iff MC(x) = AC(x) \]

2. Revenue and Profit Metrics

  • Total Revenue (TR or \(R\)): If \(p\) is unit selling price and \(x\) is quantity demanded: \[ R(x) = p \cdot x \]
  • Average Revenue (AR): \[ AR(x) = \frac{R(x)}{x} = \frac{p \cdot x}{x} = p \quad (\text{AR is always identical to unit price}) \]
  • Marginal Revenue (MR): \[ MR(x) = \frac{dR}{dx} = R'(x) \]
  • Profit Function (\(\Pi\)): \[ \Pi(x) = R(x) - C(x) \]
  • Profit Maximization Golden Rule: Profit is maximized when Marginal Revenue equals Marginal Cost: \[ \frac{d\Pi}{dx} = 0 \implies MR(x) = MC(x) \quad \text{and} \quad \frac{d^2\Pi}{dx^2} < 0 \]

3. Price Elasticity of Demand (\(\eta\))

  • Definition: The responsiveness of quantity demanded to a percentage change in price: \[ \eta = -\frac{\% \Delta x}{\% \Delta p} = -\frac{p}{x} \cdot \frac{dx}{dp} = -\frac{p / x}{\frac{dp}{dx}} \]
  • Classification:
    • \(|\eta| > 1\): Elastic Demand (A price cut increases Total Revenue).
    • \(|\eta| = 1\): Unitary Elastic Demand (Total Revenue is maximized).
    • \(|\eta| < 1\): Inelastic Demand (A price increase increases Total Revenue).
  • Amoroso-Robinson Formula: The universal relationship connecting Marginal Revenue, Average Revenue (Price), and Elasticity: \[ MR = p \left( 1 - \frac{1}{|\eta|} \right) = AR \left( 1 - \frac{1}{|\eta|} \right) \]

Comprehensive Managerial Worked Example: Monopoly Pricing and Profit Maximization

Scenario: A manufacturer faces the market demand curve \(p = 100 - 2x\), where \(p\) is unit price in ₹ and \(x\) is production volume in thousands of units. The firm's total cost function is \(C(x) = 50 + 20x + x^2\).

  1. Formulate the Total Revenue function \(R(x)\) and Marginal Revenue function \(MR(x)\).
  2. Find the Marginal Cost function \(MC(x)\).
  3. Determine the output level \(x\) that maximizes profit, and calculate the optimal selling price \(p\) and maximum profit.
  4. Compute the price elasticity of demand \(|\eta|\) at the profit-maximizing output.

Solution:

  1. Revenue: \[ R(x) = p \cdot x = (100 - 2x)x = 100x - 2x^2 \] \[ MR(x) = \frac{dR}{dx} = 100 - 4x. \]
  2. Marginal Cost: \[ MC(x) = \frac{dC}{dx} = \frac{d}{dx}(50 + 20x + x^2) = 20 + 2x. \]
  3. Profit Maximization (\(MR = MC\)): \[ 100 - 4x = 20 + 2x \implies 6x = 80 \implies x = \frac{80}{6} = \frac{40}{3} \approx 13.33 \text{ thousand units.} \] Check second order condition: \(\frac{d^2\Pi}{dx^2} = MR' - MC' = -4 - 2 = -6 < 0\) (Confirmed Maximum).

    Optimal Selling Price:

    \[ p = 100 - 2\left(\frac{40}{3}\right) = 100 - \frac{80}{3} = \frac{220}{3} \approx ₹73.33. \]

    Maximum Profit:

    \[ \Pi = R(x) - C(x) = (100x - 2x^2) - (50 + 20x + x^2) = 80x - 3x^2 - 50 \] \[ \Pi = 80\left(\frac{40}{3}\right) - 3\left(\frac{40}{3}\right)^2 - 50 = \frac{3200}{3} - \frac{1600}{3} - 50 = \frac{1600}{3} - 50 = 533.33 - 50 = ₹483.33 \text{ thousand.} \]
  4. Price Elasticity of Demand: From \(p = 100 - 2x\), we have \(\frac{dp}{dx} = -2\), so \(\frac{dx}{dp} = -\frac{1}{2}\). \[ |\eta| = \left| -\frac{p}{x} \cdot \frac{dx}{dp} \right| = \frac{220/3}{40/3} \times \frac{1}{2} = \frac{220}{40} \times \frac{1}{2} = \frac{11}{4} = 2.75. \] Verification via Amoroso-Robinson: \[ MR = p \left(1 - \frac{1}{|\eta|}\right) = \frac{220}{3} \left(1 - \frac{1}{2.75}\right) = \frac{220}{3} \left(1 - \frac{4}{11}\right) = \frac{220}{3} \times \frac{7}{11} = \frac{140}{3} \approx 46.67. \] Notice \(MC = 20 + 2(40/3) = 20 + 80/3 = 140/3\). Matches \(MR = MC\) identically!
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