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)

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