<s>[INST] 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)
 [/INST] $y$ 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,  $\max f=f(3/2).$ 
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