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 .

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 $3x+4y$.
