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The minimum value of $3x + 4y$ subject to the condition

$x^2 y^3 = 6$

and $x$ and $y$ are positive .

3 Answers 3

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Write

$3x = \frac{3x}{2} + \frac{3x}{2}$

$4y = \frac{4y}{3} + \frac{4y}{3} + \frac{4y}{3} $

and use $\text{AM} \ge \text{GM}$.

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This solution uses elementary calculus, which you may not be familiar with. Given that $x,y>0$ we can solve for $x$ in terms of $y$ as $x=\sqrt{6/y^3}$, so the problem becomes minimizing $3\sqrt{6/y^3}+4y$ with $y>0$. This can be done by taking the derivative and setting it equal to $0$: $0=\frac{d}{dy}\left(3\sqrt{6/y^3}+4y\right)=\frac{3}{2\sqrt{6/y^3}}\frac{-18}{y^4}+4=\frac{-27}{\sqrt{6y^5}}+4$ and solving, which gives us $27^2=4^2\times 6y^5$ so $y=\sqrt[5]{\frac{27^2}{4^2\times 6}}=\frac{3}{2}$ is the value of $y$ which minimizes the expression. From this you can calculate $x$ and the minimum value of the expression $3x+4y$.

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$3x+4y = 2(\frac{3x}{2})+3(\frac{4y}{3})$

According to weighted arithmetic mean and weighted geometric mean inequality,

$\left(\frac{m_1x_1 + m_2 x_2 + m_3 x_3 +...+ m_n x_n}{m_1+m_2+m_3....m_n}\right)^{m_1m_1m_2m_3.....m_n} \ge x_1^{m_1}x_2^{m_2}x_3^{m_3}x_4^{m_4}....x_n^{m_n}$

$\frac{2(\frac{3x}{2})+3(\frac{4y}{3})}{2+3} \ge[[\frac{3x}{2}]^2[\frac{4y}{3}]^3]^\frac{1}{2+3}$

$\frac{3x+4y}{5} \ge[\frac{16x^2y^3}{3}]^\frac{1}{5}$

$x^2y^3=6$

$\frac{3x+4y}{5} \ge{32}^{1/5}$

$3x+4y \ge 10$

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    Oh , ok how did I not see that before?, anyways thanks for the edit.2012-03-18