Sequences & Series
AM-GM Inequality
Grade 11

Question:

<p>If <i>p</i> and <i>q</i> are positive real numbers such that \(p^2 + q^2 = 1\), then the maximum value of \((p + q)\) is</p>
<p>2</p>
<p>\(\dfrac{1}{2}\)</p>
<p>\(\dfrac{1}{\sqrt{2}}\)</p>
<p>\(\sqrt{2}\)</p>

Step-by-Step Solution

Key Concept: Use the constraint p² + q² = 1 (circle equation) with Cauchy-Schwarz inequality: (p + q)² ≤ 2(p² + q²). Alternatively, recognize that p and q can be parameterized as p = cos θ, q = sin θ, reducing to finding max of cos θ + sin θ.
<p><strong>Step 1:</strong> Apply Cauchy-Schwarz inequality: (1·p + 1·q)² ≤ (1² + 1²)(p² + q²)</p><p><strong>Step 2:</strong> Substitute the constraint p² + q² = 1: (p + q)² ≤ 2(1) = 2</p><p><strong>Step 3:</strong> Therefore p + q ≤ √2, with equality when p/1 = q/1 and p² + q² = 1</p><p><strong>Step 4:</strong> Equality holds when p = q = 1/√2, giving maximum value = √2</p><p>∴ Answer: D (Maximum value is √2)</p>
Correct Answer: D

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