A "fast" way to ,find the maximum value of $(x^2) \times (y^3)$, if $3x+4y=12$ for $x,y \ge 0$ If  $3x+4y=12$ $\forall x,y \ge 0$, the maximum value of $(x^2)  \times (y^3)$ is


*

*$6 \times (6/5)^5$

*$3 \times (6/5)^5$

*$ (6/5)^5 $

*$7 \times (6/5)^5$


How to approach this problem? I thought of using the approach for finding maxima-minima for two independent variable but I am not sure about that as the $x$ and $y$ vanishes after the first partial derivative itself.

AM-GM gives 
$$12 = \frac{3}{2} x + \frac{3}{2} x + \frac{4}{3} y + \frac{4}{3} y + \frac{4}{3} y \ge 5 \left( \frac{16}{3} x^2 y^3 \right)^{1/5}$$
hence
$$x^2 y^3 \le \frac{3}{16} \left( \frac{12}{5} \right)^5 = 6 \cdot \left( \frac{6}{5} \right)^5$$
with equality when $\frac{3}{2} x = \frac{4}{3} y$, which is attainable.
