Home Class 11 Applied Maths Chapter 6: Sequences and Series Exercise 6.3: Relationship between AM and GM
Chapter 6 • Unit II

Exercise 6.3: Relationship between AM and GM

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

6.3 Relationship Between AM and GM: Inequality Proofs and Optimization

In Applied Mathematics, cost minimization, and financial engineering, the relationship between the Arithmetic Mean (AM) and the Geometric Mean (GM) provides one of the most powerful algebraic bounds in existence. It enables analysts to find minimum operating costs and maximum asset yields without requiring differential calculus.

Fundamental Theorems & Properties

  • The AM-GM Inequality: For any two positive real numbers \(a > 0\) and \(b > 0\): \[ \text{AM} \ge \text{GM} \iff \frac{a + b}{2} \ge \sqrt{ab} \] Equality holds (\(\text{AM} = \text{GM}\)) if and only if \(a = b\).
  • Rigorous Algebraic Proof: Consider the difference: \[ \text{AM} - \text{GM} = \frac{a + b}{2} - \sqrt{ab} = \frac{a + b - 2\sqrt{ab}}{2} = \frac{(\sqrt{a} - \sqrt{b})^2}{2} \] Since the square of any real number is non-negative, \(\frac{(\sqrt{a} - \sqrt{b})^2}{2} \ge 0\). Hence, \(\text{AM} \ge \text{GM}\).
  • Quadratic Equation Construction: If \(A\) and \(G\) are the AM and GM of two positive numbers \(a\) and \(b\): \[ a + b = 2A, \quad ab = G^2 \] The quadratic equation whose roots are \(a\) and \(b\) is: \[ x^2 - 2Ax + G^2 = 0 \] By the quadratic formula, the numbers are given by: \[ a, b = A \pm \sqrt{A^2 - G^2} \]
  • The Ratio Theorem: If the AM of two positive numbers is to their GM in the ratio \(m : n\) (where \(m > n > 0\)), then the ratio of the two numbers is: \[ \frac{a}{b} = \frac{m + \sqrt{m^2 - n^2}}{m - \sqrt{m^2 - n^2}} \]
  • Generalization to \(n\) Positive Variables: \[ \frac{x_1 + x_2 + \dots + x_n}{n} \ge \sqrt[n]{x_1 x_2 \dots x_n} \]

Applied Business Optimization: Minimizing Operating Cost

Example: Inventory Carrying and Ordering Costs

Scenario: A supply chain manager models the annual operating cost \(C(x)\) of maintaining a warehouse stocking quantity \(x > 0\) as: \[ C(x) = 4x + \frac{2500}{x} \] where \(4x\) represents carrying storage costs and \(\frac{2500}{x}\) represents ordering overheads. Using the AM-GM inequality, determine the minimum annual operating cost and the optimal stocking quantity \(x\).

Solution:

Let \(a = 4x\) and \(b = \frac{2500}{x}\). Since \(x > 0\), both \(a > 0\) and \(b > 0\).

By the AM-GM Inequality:

\[ \frac{a + b}{2} \ge \sqrt{ab} \implies a + b \ge 2\sqrt{ab} \] \[ C(x) = 4x + \frac{2500}{x} \ge 2 \sqrt{4x \cdot \frac{2500}{x}} = 2 \sqrt{10{,}000} = 2 \times 100 = 200 \]

The minimum annual operating cost is ₹200.

This minimum occurs when equality holds in the AM-GM inequality, which requires \(a = b\):

\[ 4x = \frac{2500}{x} \implies 4x^2 = 2500 \implies x^2 = 625 \implies x = 25 \text{ units.} \]

Thus, the optimal stocking order is 25 units.

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