Let x , y > 0 . If x 3 y 2 = 2 15 , then the least value of 3 x + 2 y is

Using $A.M\ge G.M$, we get

$\frac{x+x+x+y+y}{5}\ge \sqrt[5]{x^3·{\left(y\right)}^2}$

$\Rightarrow \frac{3x+2y}{5}\ge \sqrt[5]{x^3{y}^2}$

$\Rightarrow \frac{3x+2y}{5}\ge \sqrt[5]{\left(2^{15}\right)} \left(\because x^3{y}^2=2^{15}\right)$

$\Rightarrow 3x+2y\ge 5·{\left(2\right)}^{15/5}$

$\Rightarrow 3x+2y⩾5\times 2^3$

$\Rightarrow 3x+2y⩾40$

$\therefore$   The minimum value of $3x+2y$ is $40$