If 2x+3y=12; x, y $>$ 0, then the maximum possible value of $x y^{2}$ is ___.

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Solution

Power of x is 1 and that of y is 2 in $x y^{2} .$
So, we split the term with y into two.
2x+2y+y=12
He know that, AM $\geq$ GM.
$\frac{2 x + 2 y + y}{3} \geq (2 x \times 2 y \times y)^{\frac{1}{3}}$

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