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.
 A: AM-GM gives 
$$12 = \frac{3}{2} x + \frac{3}{2} x + \frac{4}{3} y + \frac{4}{3} y + \frac{4}{3} y \ge 5 \left( \frac{16}{3} x^2 y^3 \right)^{1/5}$$
hence
$$x^2 y^3 \le \frac{3}{16} \left( \frac{12}{5} \right)^5 = 6 \cdot \left( \frac{6}{5} \right)^5$$
with equality when $\frac{3}{2} x = \frac{4}{3} y$, which is attainable.
A: 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 $(8/5)^2(9/5)^3$. 
A: Here's a bit of a tricky way to do it: instead of maximizing $x^2 y^3$, we'll maximize its logarithm $2 \log x + 3 \log y$, or alternatively $2 \log x + 3 \log (3 - 3x/4)$. 
But that denominator of $4$ is annoying; since
$$ \log (3 - 3x/4) = \log (12-3x) - \log 4 $$
we can minimize
$$ f(x) = 2 \log x + 3 \log (12-3x) $$
instead.
Differentiate: $f^\prime(x) = 2/x - 9/(12-3x)$. This is zero when $2(12-3x) = 9x$, or $24-6x = 9x$, or $x = 8/5$. Since $3x+4y=12$ we can solve to get $y = 9/5$. The maximum value of $x^2 y^3$ is therefore 
$$ (8/5)^2 (9/5)^3 = (4/3 \times 6/5)^2 (3/2 \times 6/5)^3  = (4/3)^2 (3/2)^3 (6/5)^5 = 6 (6/5)^5. $$
This is probably not the most efficient solution to this particular problem.  (This is made worse by the fact that once you get the answer, it still takes a while to figure out which of the choices it is; I suspect it's not what the writers of whatever test you're preparing for had in mind.) But I give it because it illustrates a general principle: products are annoying, so if you want to maximize one sometimes it's easier to maximize its log, which is a sum.
