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


$$x^2 y^3  = 6$$ 


and $x$ and $y$ are positive .
 A: Write
$$3x = \frac{3x}{2} + \frac{3x}{2}$$
$$4y = \frac{4y}{3} + \frac{4y}{3}  + \frac{4y}{3} $$
and use $\text{AM} \ge \text{GM}$.
A: 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 $3x+4y$.
A: $$3x+4y = 2(\frac{3x}{2})+3(\frac{4y}{3})$$
According to weighted arithmetic mean and weighted geometric mean inequality,

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

$$\frac{2(\frac{3x}{2})+3(\frac{4y}{3})}{2+3} \ge[[\frac{3x}{2}]^2[\frac{4y}{3}]^3]^\frac{1}{2+3}$$
$$\frac{3x+4y}{5} \ge[\frac{16x^2y^3}{3}]^\frac{1}{5}$$
$x^2y^3=6$
$$\frac{3x+4y}{5} \ge{32}^{1/5}$$
$$3x+4y \ge 10$$
