Question

The greatest value of $\mathrm{xyz}$ for $+\mathrm{ve}$ values of $\mathrm{x},\mathrm{y},\mathrm{z}$ ; where $\mathrm{yz}+\mathrm{zx}+\mathrm{xy}=12$ is

• A. $4$
• B. $6$
• C. $8$
• D. $10$

Answer 1

The correct option is C

$8$

The explanation for the correct answer.

Find the greatest value.

Given: $\mathrm{yz}+\mathrm{zx}+\mathrm{xy}=12$

Using the relation AM $\ge$GM, AM is Arithmetic Mean and GM is Geometric Mean.

$$\begin{gathered} \frac{xy+yx+zx}{3}\ge {\left(x^2{y}^2{z}^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ \Rightarrow 4\ge {\left(xyz\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ \Rightarrow 4^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\ge xyz \\ \Rightarrow 8\ge xyz \end{gathered}$$

Hence option $\left(c\right)$ is the correct answer.