Let $x, y, z > 0$ with $x + y + z = 12$ and $x^3y^4z^5 = (0.1) \times 600^3$. Then $x^3 + y^3 + z^3$ equals:
Step-by-Step Solution
Key Concept: The weighted AM-GM for $x + y + z = 12$ with weights $3, 4, 5$ gives maximum of $x^3y^4z^5$ when $x/3 = y/4 = z/5 = 12/12 = 1$, i.e. $x = 3, y = 4, z = 5$. Verify the product equals $(0.1) \cdot 600^3$, then compute the sum of cubes.
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Correct Answer: (2)