A "fast" way to ,find the maximum value of $(x^2) \times (y^3)$, if $3x+4y=12$ for $x,y \ge 0$ If  $3x+4y=12$ $\forall x,y \ge 0$, the maximum value of $(x^2)  \times (y^3)$ is


*

*$6 \times (6/5)^5$

*$3 \times (6/5)^5$

*$ (6/5)^5 $

*$7 \times (6/5)^5$


How to approach this problem? I thought of using the approach for finding maxima-minima for two independent variable but I am not sure about that as the $x$ and $y$ vanishes after the first partial derivative itself.

Lagrange multipliers are pretty fast, and don't require seeing any tricks:
$$2xy^3 = 3\lambda$$
$$3x^2y^2 = 4\lambda$$
Divide the two equations, and you get ${\displaystyle {2 \over 3} {y \over x} = {3 \over 4}}$ or ${\displaystyle y = {9 \over 8} x}$. Putting back into the equation you get
$$3x + {9 \over 2} x = 12$$
This solves as $x = 24/15 = 8/5$ and $y = 9/8\times 8/5 = 9/5$. So the minimum value of $x^2y^3$ is $(8/5)^2(9/5)^3$. 
