Question

Let $x, y ,z $ be positive real numbers such that $x+y+z=12$ and $x^3 y^4 z^5 = (0.1) (600)^3$. Then $x^3 + y^3 + z^3$ is equal to :

• A. 270
• B. 258
• C. 342
• D. 216

Answer 1

Answer: (D)

$x+y+z=12$
$AM \ge GM$
$\frac{3\left(\frac{x}{3}\right)+4\left(\frac{y}{4}\right)+5\left(\frac{z}{5}\right)}{12} \ge\sqrt[12]{\left(\frac{x}{3}\right)^{3}\left(\frac{y}{4}\right)^{2}\left(\frac{z}{5}\right)^{5}}$
$\frac{x^{3}y^{4}z^{5}}{3^{3}4^{4}5^{5}}\le1$
$x^{3}y^{4}z^{5} \le3^{3},4^{4},5^{5}$
$x^{3}y^{4}z^{5} \le \left(0.1\right)\left(600\right)^{3}$
But, given $x^{3}y^{4}z^{5}=\left(0.1\right)\left(600\right)^{3}$
$\therefore $ equality occurs
$\therefore \frac{x}{3}=\frac{y}{4}=\frac{z}{5}\left(=k\right)$
$x=3k; y=4k; z=5k$
$x+y+z=12$
$3k + 4k + 5y =12$
$k=1M$
$\therefore x=3; y=4; z=5$
$\therefore x^{3}+y^{3}+z^{3}=216$