<s>[INST] range of $3x^2-2xy$ subjected to $x^2+y^2=1$ If $x^2+y^2=1$. then the range of expression $3x^2-2xy$ without trigonometric substitution method
what i have done  try here is  use arithmetic geometric inequality
$\displaystyle x^2+y^2\geq 2xy$
$\displaystyle -2xy\geq -(x^2+y^2)$
$\displaystyle 3x^2-2xy\geq 2x^2-y^2$
this will not help more
how do i solve it help  me please
 [/INST] Since $x^2+y^2=1$, you can substitute for $y=\pm\sqrt{1-x^2}$ in $f(x,y)=3x^2-2xy$ to get $g(x)=f(x,\pm\sqrt{1-x^2})=3x^2\pm2x\sqrt{1-x^2},-1\le x\le1$ and find the extrema using standard techniques.
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