Inequality based on AM/GM Inequality Find the greatest value of $x^3y^4$
If $2x+3y=7 $ and $x≥0, y≥0$.
(Probably based on weighted arithmetic and geometric mean)
 A: Use this Hint:
$$7=\frac23x+\frac23x+\frac23x + \frac34y+\frac34y+\frac34y+\frac34y\geqslant 7\sqrt[7]{\cdots}$$
A: it's pretty simple if you know about weighted means, proceed like this:
$(2x/3)*3+(3y/4)*4 ≥ 7[(2x/3)^³(3y/4)^4)^{1/7}$
i.e. $7≥7((x^3y^4)(3/32))^{1/7}$
or, $(x^3y^4)(3/32)≤1$
hence, $x^3y^4≤32/3$
so $32/3$ is the greatest value of $x^3y^4$
PS: sorry for the ill written answer, I'm on phone too, my friends out there please edit the answer ;)
A: My usual attempt to generalize.
To find the max of
$x^m y^n$
subject to
$ax+by = c$
where
$m$ and  $n$
are integers.
Following
Macavity's solution,
use the
AGM inequality
in the form
$\sum_{i=1}^p a_i
\ge p (\prod_{i=1}^p a_i)^{1/p}
$.
Then
$\begin{array}\\
c
&=m\frac{ax}{m}
+n\frac{by}{n}\\
&\ge (m+n)((\frac{ax}{m})^m(\frac{by}{n})^n)^{1/(m+n)}
\qquad (m \text{ copies of }\frac{ax}{m}
\text{ and }n\text{ copies of }\frac{by}{n})\\
\end{array}
$
with equality,
by the AGM inequality,
if and only if
$\frac{ax}{m}
=\frac{by}{n}
$.
If this is true,
then
$y
=\frac{anx}{bm}
$
so
$\begin{array}\\
c
&=m\frac{ax}{m}
+n\frac{by}{n}\\
&=m\frac{ax}{m}
+n\frac{ax}{m}\\
&=ax(1+\frac{n}{m})\\
&=ax\frac{m+n}{m}\\
\end{array}
$
or
$x
=\frac{mc}{a(m+n)}
$
and
$y
=\frac{mc}{a(m+n)}\frac{an}{bm}
=\frac{nc}{b(m+n)}
$.
Finally,
$x^m y^n
=(\frac{mc}{a(m+n)})^m(\frac{nc}{b(m+n)})^n
=(\frac{m}{a})^m(\frac{n}{b})^n(\frac{c}{m+n})^{m+n}
$.
If $m+n = c$,
as in OP's case,
$x
=\frac{m}{a}
$
and
$y
=\frac{n}{b}
$
and the max is
$x^m y^n
=(\frac{m}{a})^m(\frac{n}{b})^n
$.
In this case,
$m=3, n=4, a=2, b=3, c=7$.
The result is
$x 
= \frac{3}{2}
$,
$y 
= \frac{4}{3}
$,
and the max is
$(\frac{3}{2})^3(\frac{4}{3})^4
=\frac{2^5}{3}
$.
Note that
this generalizes easily to
maximize
$\prod_{i=1}^m x_i^{n_i}
$
subject to
$c
=\sum_{i=1}^m a_ix_i
$
with
$c, n_i,$
and $a_i$
being specified.
A: $y$ is a function of $x$. That is, $y=(7-2x)/3.$ And $\frac {dy}{dx}=y'=-2/3.$ 
Let $f(x)=x^3y^4.$ Then $f'(x)=3x^2 y^4+4x^3y^3y'=3x^2y^4-(8/3)x^3y^3.$
For $xy\ne 0$ we have $f(x)>0$ and $f'(x)=f(x)(\frac {3}{x}-\frac {8}{3y}).$
So for $xy\ne 0$ we have $f'(x)>0\iff \frac {3}{x}-\frac {8}{3y}>0\iff 9y>8x\iff21-6x>8x \iff 3/2>x.$ Similarly for $xy\ne 0$ we have $f'(x)<0\iff 3/2<x.$
And we have $xy=0\implies f'(x)=0.$
So $f(x)$ is increasing for $x<3/2$ and is decreasing for $3/2<x<7/2.$ ($7/2$ is the upper limit of the domain of $f,$ as $y=(7-2x)/3\geq 0.$) Therefore, since $f(x)$ is continuous,  $\max f=f(3/2).$ 
