Question

If $x+y=8$, then the maximum value of $x^{2}y$ is:

• A. $\frac{2048}{9}$
• B. $\frac{2048}{81}$
• C. $\frac{2048}{3}$
• D. $\frac{2048}{27}$

Answer 1

The correct option is D

$\frac{2048}{27}$

Explanation for the correct option.

In $x^{2}y$, the power of x is $2$, so we will divide it into two parts, that is $\frac{x}{2},\frac{x}{2}$.

Now, the arithmetic mean of $\frac{x}{2},\frac{x}{2},y$ will be $\frac{{\displaystyle \frac{x}{2}}+{\displaystyle \frac{x}{2}}+y}{3}$, which will be $\frac{x+y}{3}$.

The geometric mean of $\frac{x}{2},\frac{x}{2},y$ will be ${\left(\frac{x}{2}\times \frac{x}{2}\times y\right)}^{1/3}$, which will be ${\left(\frac{x^{2}y}{4}\right)}^{1/3}$.

$\begin{array}{rcl}& \Rightarrow & \frac{x+y}{3}\ge {\left(\frac{x^{2}y}{4}\right)}^{1/3}\left[AsA.M\ge G.M\right]\\ \Rightarrow \frac{8}{3}& \ge & {\left(\frac{x^{2}y}{4}\right)}^{1/3}\end{array}$

Taking cube on both the sides, we get

$\begin{array}{rcl}\frac{x^{2}y}{4}& \le & \frac{512}{27}\\ & \Rightarrow & x^{2}y\le \frac{2048}{27}\end{array}$

Hence, option D is correct.