A) 270
B) 258
C) 216
D) 342
Correct Answer: C
Solution :
\[\text{x}+\text{y}+\text{z}=\text{12}\] \[{{x}^{3}}{{y}^{4}}{{z}^{5}}=(0.1){{(600)}^{3}}\] \[\frac{3\left( \frac{x}{3} \right)+4\left( \frac{y}{4} \right)+5\left( \frac{z}{5} \right)}{12}\ge {{\left\{ {{\left( \frac{x}{3} \right)}^{3}}{{\left( \frac{y}{4} \right)}^{4}}{{\left( \frac{z}{5} \right)}^{5}} \right\}}^{1/12}}\] \[1\ge \frac{{{x}^{3}}{{y}^{4}}{{z}^{5}}}{{{(60)}^{3}}(4\times 25)}\] \[{{x}^{3}}{{y}^{4}}{{z}^{5}}\le (0.1){{(600)}^{3}}\] But \[{{x}^{3}}{{y}^{4}}{{z}^{5}}=(10.1){{(600)}^{3}}\] Clearly AM = GM Hence\[\frac{x}{3}=\frac{y}{4}=\frac{z}{3}\] \[\Rightarrow \]\[x=3,y=4,z=5\]\[\Rightarrow \] \[\Rightarrow \]\[{{x}^{3}}+{{y}^{3}}+{{z}^{3}}=27+64+125=216\]
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