Single correct option
If x and y are positive real numbers with x + y = 12, the maximum possible value of x²y is
- A. 216
- B. 243
- C. 256
- D. 288
Show answer and solution
Answer: (C) 256
Write x + y = x/2 + x/2 + y = 12. By AM ≥ GM on the three terms x/2, x/2, y:
12/3 ≥ ∛((x/2)(x/2)y) → 64 ≥ x²y/4 → x²y ≤ 256.
Equality holds when x/2 = y, i.e. x = 8, y = 4: 64 × 4 = 256.
Why other options fail: 216 = 6³ comes from wrongly equating x = y = 6; 288 cannot be attained.
12/3 ≥ ∛((x/2)(x/2)y) → 64 ≥ x²y/4 → x²y ≤ 256.
Equality holds when x/2 = y, i.e. x = 8, y = 4: 64 × 4 = 256.
Why other options fail: 216 = 6³ comes from wrongly equating x = y = 6; 288 cannot be attained.
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