Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>If \(a > 0\), \(b > 0\), \(c > 0\) and \(2a + b + 3c = 1\), then which of the following is correct?</p>
<p>(1) \(a^4 b^2 c^2\) is greatest then \(a = \dfrac{1}{4}\)</p>
<p>(2) \(a^4 b^2 c^2\) is greatest then \(b = \dfrac{1}{4}\)</p>
<p>(3) \(a^4 b^2 c^2\) is greatest then \(c = \dfrac{1}{12}\)</p>
<p>(4) greatest value of \(a^4 b^2 c^2\) is \(\dfrac{1}{9 \cdot 4^8}\)</p>
Step-by-Step Solution
Key Concept: Use AM-GM inequality on weighted terms or apply Cauchy-Schwarz after rewriting the constraint. The constraint 2a + b + 3c = 1 suggests applying weighted AM-GM or recognizing this as an optimization problem where the minimum of a symmetric expression occurs when terms are proportional to their coefficients.
<p><strong>Step 1:</strong> Given constraint: 2a + b + 3c = 1 with a, b, c > 0.</p><p><strong>Step 2:</strong> By Cauchy-Schwarz inequality: (2a + b + 3c)(1/2a + 1/b + 1/3c) ≥ (√2·1/√2 + √1·1/√1 + √3·1/√3)² = (1 + 1 + 1)² = 9</p><p><strong>Step 3:</strong> Since 2a + b + 3c = 1, we have: 1/2a + 1/b + 1/3c ≥ 9</p><p><strong>Step 4:</strong> Equality holds when 2a/(1/2a) = b/(1/b) = 3c/(1/3c), which gives 4a² = b² = 9c², so a:b:c = 1:2:3 (with proportional scaling).</p><p><strong>Step 5:</strong> From 2a + b + 3c = 1 with a = k, b = 2k, c = 3k: 2k + 2k + 9k = 1, giving k = 1/13. Thus minimum of 1/2a + 1/b + 1/3c is 9.</p><p>∴ Answer: C</p>
Correct Answer: C