Given that $x+y=6$. Is there any way to find the maximum value of $x^2y$ without using calculus or graphical method?


Does not exist! Try $y\rightarrow+\infty$.

For non-negative variables we can use AM-GM: $$x^2y=\frac{1}{2}x\cdot x\cdot2y\leq\frac{1}{2}\left(\frac{x+x+2y}{3}\right)^3=32.$$ The equality occurs for $x=4$ and $y=2$, which says that $32$ is a maximal value.

  • $\begingroup$ Can I know what your meaning of does not exist? $\endgroup$ – Mathxx Aug 11 '17 at 9:12
  • 2
    $\begingroup$ @Mathxx It means $x^2y\rightarrow+\infty$ for $y\rightarrow+\infty$. $\endgroup$ – Michael Rozenberg Aug 11 '17 at 9:16
  • $\begingroup$ He means that you need to add that $x$ and $y$ are non negative to your hypothesis. $\endgroup$ – Joaquin San Aug 11 '17 at 9:35
  • $\begingroup$ @MichaelRozenberg How do you know the equality happens at that value? $\endgroup$ – Mathxx Aug 11 '17 at 11:43
  • $\begingroup$ @Mathxx Substitute $x=4$ and $y=2$: $x+y=6$ and $x^2y=4^2\cdot2=32$. $\endgroup$ – Michael Rozenberg Aug 11 '17 at 11:47

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