Basic Mathematics & Logarithm
Logarithmic Inequalities
Grade 11
Question:
<p>If \(\log_{10}(x^3 + y^3) - \log_{10}(x^2 + y^2 - xy) \leq 2\), where \(x, y\) are positive real numbers, then find the maximum value of \(xy\).</p>
Step-by-Step Solution
Key Concept: Recognize that x³ + y³ = (x + y)(x² + y² - xy), so the logarithmic inequality simplifies to log₁₀(x + y) ≤ 2, which gives x + y ≤ 100. Then apply AM-GM inequality to maximize xy subject to this constraint.
<p><strong>Step 1:</strong> Use the factorization identity: x³ + y³ = (x + y)(x² + y² - xy)</p><p><strong>Step 2:</strong> Simplify the logarithmic inequality:</p><p>log₁₀[(x + y)(x² + y² - xy)] - log₁₀(x² + y² - xy) ≤ 2</p><p>log₁₀[(x + y)(x² + y² - xy)/(x² + y² - xy)] ≤ 2</p><p>log₁₀(x + y) ≤ 2</p><p><strong>Step 3:</strong> Convert from logarithmic form:</p><p>x + y ≤ 10² = 100</p><p><strong>Step 4:</strong> Apply AM-GM inequality to maximize xy:</p><p>(x + y)/2 ≥ √(xy)</p><p>xy ≤ [(x + y)/2]²</p><p><strong>Step 5:</strong> Equality holds when x = y. With x + y = 100 (maximum constraint):</p><p>x = y = 50</p><p>xy = 50 × 50 = 2500</p><p><strong>Verification:</strong> When x = y = 50: log₁₀(125000) - log₁₀(2500) = log₁₀(50) = log₁₀(10²/2) ≈ 1.699 < 2 ✓</p><p>∴ Maximum value of xy = <strong>2500</strong></p>
Correct Answer: 2500