Question

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

• A. $342$
• B. $342$
• C. $216$
• D. $270$

Answer 1

The correct option is C $216$

Relation between $AM,GMandHM$ of two positive numbers -

$AM\ge GM\ge HM$

wherein

Inequality of the three given means.

We have weighted $A.M\ge$ weighted $G.M$.

$⟹\frac{3\left(\frac{x}{3}\right)+4\left(\frac{y}{4}\right)+5\left(\frac{z}{5}\right)}{12}\ge \left(\right(\frac{x}{3}{)}^3\left(\frac{y}{4}{)}^4\right(\frac{z}{5}{)}^5{)}^{\frac{1}{12}}$

$1\ge {\frac{x^3{y}^4{2}^5}{3^3{4}^4{5}^5}}^{\frac{1}{12}}⟹\left(x^3{y}^4{z}^5\right)\le \left(0.1\right)(600{)}^3$

But $x^3{y}^4{z}^5=\left(0.1\right)(600{)}^3$

Which means as given $A.M=G.M$

$⟹\frac{x}{3}=\frac{y}{4}=\frac{z}{5}$

$⟹x^3+y^3+z^3=3^3+4^3+5^3=216$