<p>Let <em>x</em>, <em>y</em>, <em>z</em> be positive real numbers such that \(x+y+z=12\) and \(x^3y^4z^5=(0.1)(600)^3\). Then \(x^3+y^3+z^3\) is equal to</p>
Step-by-Step Solution
Key Concept: Use weighted AM-GM inequality on the constraint x³y⁴z⁵ to find optimal values of x, y, z, then apply Lagrange multipliers or recognize that the maximum of the product occurs when the variables are proportional to their exponents.
<p><strong>Step 1:</strong> From the constraint x³y⁴z⁵ = (0.1)(600)³ = 60³/10 = 216000, we need to find x, y, z that satisfy both x+y+z=12 and this product condition.</p><p><strong>Step 2:</strong> Apply AM-GM with weights 3, 4, 5 (total weight = 12). For the product x³y⁴z⁵ to achieve its maximum given x+y+z=12, we need:</p><p>$$\frac{3x + 4y + 5z}{12} = \frac{x+y+z}{1}$$</p><p>This gives us x/3 = y/4 = z/5 = k for some constant k.</p><p><strong>Step 3:</strong> From x+y+z=12: 3k + 4k + 5k = 12, so 12k = 12, thus k = 1.</p><p>Therefore: x = 3, y = 4, z = 5</p><p><strong>Step 4:</strong> Verify: x³y⁴z⁵ = 3³·4⁴·5⁵ = 27·256·3125 = 216000 ✓</p><p><strong>Step 5:</strong> Calculate x³+y³+z³ = 3³+4³+5³ = 27+64+125 = 216</p><p>∴ Answer: D (216)</p>
Correct Answer: D