<s>[INST] Maximum of function containing two variables $x$ and $y$ 
If $x+y+\sqrt{2x^2+2xy+3y^2} = k(\bf{Const.})\;,$ Then $\max(x^2y)\;,$ Where $x,y\geq 0$

$\bf{My\; Try::}$ Let $x^2y=z\;$ Then we get $$x+\frac{x^2}{z}+\sqrt{2x^2+\frac{2z}{x}+\frac{3z^2}{x^4}} = k$$
Now How can I solve it, Help me
Thanks
 [/INST] Assuming $y=a x$ ($a\geq0$), the constraint leads to $$x=\frac{k}{\sqrt{3 a^2+2 a+2}+a+1}$$ and $$x^2y=\frac{a k^3}{\left(\sqrt{3 a^2+2 a+2}+a+1\right)^3}$$ Now, computing $$\frac{d}{da}x^2y=-\frac{(2 a-1) \left(3 a+\sqrt{a (3 a+2)+2}+2\right)}{\sqrt{a (3 a+2)+2}
   \left(a+\sqrt{a (3 a+2)+2}+1\right)^4}k^3$$ which cancels if $a=\frac 12$.
All of this makes a maximum value equal to $$\frac{4 k^3}{\left(3+\sqrt{15}\right)^3}$$
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