If $x^{2} y^{3} = 6,$ find the least value of $3 x + 4 y$ for postivie value of x and y

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Solution

Given if $x^{2} y^{3} = 6$
we have to find last value of $3 x + 4 y$

Break down $3 x$ into $3 / 2, 3 x / 2$ and $4 y$ into $4 y / 3, 4 y / 3$
$\frac{4 y}{3}$

$A . M$ of these division $\geq G . M$ of these division

$\frac{\frac{3 x}{2} + \frac{3 x}{2} + \frac{4 y}{3} + \frac{4 y}{3} + \frac{4 y}{3}}{5} \geq {(\frac{3}{2} \times \frac{3}{2} \times \frac{4}{3} \times \frac{4}{3} \times \frac{4}{3})}^{1 / 5}$

$\Rightarrow \frac{3 x + 4 y}{5} \geq (2^{5})^{1 / 5}$

$\Rightarrow 3 x + 4 y \geq 10$

Minimum value $= 10$

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Similar questions

x^{2} . y^{3}

x = y^{3} + 4 y

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∠ B = (4 y + 6),

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