Find the maximum value of $x^2y^3$ subject to the condition that $3x+2y=1$

Question

I am trying to use G.M.$\leq $ A.M.as follows

$(\frac{2x^2y^3}{4})^\frac{1}{5} \leq \frac{x+2x+y+\frac{y}{2} +\frac{y}{2}}{5}$

$\implies (\frac{x^2y^3}{2})^\frac{1}{5} \leq \frac{1}{5}$

$\implies$ $x^2y^3\leq \frac{2}{5^5}$.

Is it correct ?

Answer 1

For positive $x$ and $y$, you may use AM-GM:

$$\frac{1}{5}=\frac{3x+2y}{5}=\frac{\frac{3x}{2}+\frac{3x}{2}+\frac{2y}{3}+\frac{2y}{3}+\frac{2y}{3}}{5}\ge \sqrt[5]{\frac{2}{3}x^2y^3}$$

From where one gets $x^2y^3\le\frac{3}{2\cdot 5^5}=\frac{3}{6250}$

The equality takes place when $\frac{3x}{2}=\frac{2y}{3}$, which coupled with $3x+2y=1$ yields $x=\frac{2}{15}$ and $y=\frac{3}{10}$

But of course, the general method (not precalculus level as tagged) to solve this optimization problem is by using the method of Lagrange Multipliers.

As others noted, if we omit $x,y>0$ there is no global maximum.