The minimum value of  $3x+4y \text{for }x>0,y>0$  such that  $x^2{y}^3=6$  is

$3x+4y=\frac{3x}{2}+\frac{3x}{2}+\frac{4y}{3}+\frac{4y}{3}+\frac{4y}{3}$

$\ge 5{(\frac{3x}{2}\cdot \frac{3x}{2}\cdot \frac{4y}{3}\cdot \frac{4y}{3}\cdot \frac{4y}{3})}^{1/5}=5{\left(\frac{16}{3}{x}^2{y}^3\right)}^{\frac{1}{5}}$

$=5{\left(\frac{16}{3}.3\right)}^{\frac{1}{5}}=5(32{)}^{\frac{1}{5}}=5\times 2=10$

$\therefore$ The minimum value of  $3x+4y$ is 10