Question

Find the maximum value of $3x+4y$ for $x>0,y>0$ such that $x^2{y}^3=6$.

Answer 1

Let $z=3x+4y$

$3x+4y$ is the given equation

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

$y^3=\frac{6}{x^2}$ or $x^2=\frac{6}{y^3}$

$x=\frac{\sqrt{6}}{y^{3/2}}$

$z=\frac{3\sqrt{6}}{y^{3/2}}+4y$

We have to minimum $z$

$z=f\left(y\right)$

$f^{\prime }\left(y\right)=\frac{d}{dy}(\frac{3\sqrt{6}}{y^{3/2}}+4y)$

$=\frac{3\sqrt{6}}{y^{5/2}}\left(\frac{-3}{2}\right)+4$

Put $f^{\prime }\left(y\right)=0$ i.e $y=\frac{9\sqrt{6}}{2y^{5/2}}=0$

$=y=\frac{9\sqrt{6}}{y^{5/2}}{y}^{5/2}=\frac{9\sqrt{6}}{8}$

$=y^3=\frac{91\times 6}{64}=\frac{243}{32}$

Thus $f\left(y\right)$ is max at $y=\frac{3}{2}=\frac{3\sqrt{6}}{3\sqrt{3}}-2\sqrt{2}+4-\frac{3}{2}=4+6=10$