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$3x+4y=10$

$x>0,y>0$

$(x^2y^3)_{max}=?$

For positive numbers, A.M. $\geq $ G.M.

So, $\frac{{\frac{{3x}}{2} + \frac{{3x}}{2} + \frac{{4y}}{3} + \frac{{4y}}{3} + \frac{{4y}}{3}}}{5} \ge {\left[ {{{\left( {\frac{{3x}}{2}} \right)}^2}{{\left( {\frac{{4y}}{3}} \right)}^3}} \right]^{1/5}}$

$ \Rightarrow \frac{{3x + 4y}}{5} \ge {\left( {\frac{9}{4}{x^2} \times \frac{{4 \times 16}}{{9 \times 3}}{y^3}} \right)^{1/5}}$

$ \Rightarrow \frac{{10}}{5} \ge {\left( {\frac{{16}}{3}{x^2}{y^3}} \right)^{1/5}}$

$ \Rightarrow {2^5} \ge \frac{{16}}{3}{x^2}{y^3}$

$ \Rightarrow {x^2}{y^3} \le 6$

$ \Rightarrow {\left( {{x^2}{y^3}} \right)_{\max }} = 6$

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