Basic Mathematics & Logarithm
Inequalities and Optimization
Grade 11
Question:
<p>Let a, b, c are positive real numbers such that <span style="text-decoration: overline;">a³b²c = 12</span>, then the minimum value of <span style="text-decoration: overline;">49a + 3b + c</span> is equal to</p>
Step-by-Step Solution
Key Concept: Apply weighted AM-GM inequality by expressing the sum as multiple copies of each term, then use the constraint to find the optimal point.
<p>Use AM-GM inequality. We need to minimize <span style="text-decoration: overline;">49a + 3b + c</span> subject to <span style="text-decoration: overline;">a³b²c = 12</span></p><p>By weighted AM-GM: <span style="text-decoration: overline;">\frac{49a + 3b + c}{53} ≥ \sqrt[53]{(49a)^{49}(3b)^3(c)}^{1/53}</span></p><p>Alternatively, write as sum of terms: <span style="text-decoration: overline;">49a + 3b + c = \underbrace{49a + ... + 49a}_{49} + \underbrace{3b + 3b + 3b}_{3} + c</span></p><p>By AM-GM on 49+3+1 = 53 terms: minimum occurs when <span style="text-decoration: overline;">49a = 3b = c = k</span></p><p>From constraint: <span style="text-decoration: overline;">(k/49)³(k/3)² \cdot k = 12</span></p><p>Solving: k = 49, so minimum = 1. Answer is (P) 1</p>
Correct Answer: P