Question

The greatest value of $x^2{y}^3$ where $x>0,y>0$ and $3x+4y=5$ is

• A. $\frac{3}{8}$
• B. $\frac{3}{16}$
• C. $\frac{6}{5}$
• D. $\frac{9}{5}$

Answer 1

The correct option is B $\frac{3}{16}$

Since, $x>0,y>0$ so using AM-GM inequality, we have
$\frac{2\cdot \frac{3x}{2}+3\cdot \frac{4y}{3}}{5}\ge {\left({\left(\frac{3x}{2}\right)}^2{\left(\frac{4y}{3}\right)}^3\right)}^{1/5}$
$\Rightarrow x^2{y}^3\le \frac{3}{16}$