Find the greatest value of $x^{2} y^{3}$ subject to the conditions $3 x + 4 y = 5,$ where x and y are positive numbers.

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Solution

$\frac{x + x + y + y + y}{5} \geq {[x^{2} y^{3}]}^{1 / 5} 3 x + 4 y = 5 \frac{x + 2 x + y + y + 2 y}{5} \geq {[2 x^{2} 2 y^{3}]}^{1 / 5} . \frac{5}{5} \geq 2^{2 / 5} {[x^{2} y^{3}]}^{1 / 5} \frac{1}{2^{2 / 5}} \geq {[x^{2} y^{3}]}^{1 / 5} \frac{1}{2^{2}} \geq x^{2} y^{3} \frac{1}{4} \geq x^{2} y^{3}$

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