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
On
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.
On
The range that you're after is $\left[\frac{3-\sqrt{13}}2,\frac{3+\sqrt{13}}2\right]$. In fact, if $x^2+y^2=1$, then$$3x^2-2xy=3x^2\pm x\sqrt{1-x^2}.$$So, for each $x\in[-1,1]$, let$$f(x)=3x^2-2x\sqrt{1-x^2}\text{ and let }g(x)=3x^2+x\sqrt{1-x^2}.$$Now, using that standard Calculus techniques, you can check that the range of both functions $f$ and $g$ is $\left[\frac{3-\sqrt{13}}2,\frac{3+\sqrt{13}}2\right]$.
On
For a calculs-free approach, note that $$ \frac{3+\sqrt{13}}{2}\left[\frac{2y-(3+\sqrt{13})x}{\sqrt{26+6\sqrt{13}}}\right]^2+\frac{3-\sqrt{13}}{2}\left[\frac{2y-(3-\sqrt{13})x}{\sqrt{26-6\sqrt{13}}}\right]^2=3x^2-2xy $$ and $$ \left[\frac{2y-(3+\sqrt{13})x}{\sqrt{26+6\sqrt{13}}}\right]^2+\left[\frac{2y-(3-\sqrt{13})x}{\sqrt{26-6\sqrt{13}}}\right]^2=x^2+y^2 $$ So the range of $3x^2-2xy$, as $(x,y)$ varies on the unit circle, is the closed interval with endpoints $\dfrac{3\pm\sqrt{13}}{2}$.
On
If you have the requisite linear algebra theorems at your disposal, then you can recognize
$$3x^2-2xy=\pmatrix{x&y}\pmatrix{3&-1\\-1&0}\pmatrix{x\\y}$$
where the $2\times2$ matrix is selfadjoint with eigenvalues satisfying $\lambda^2-3\lambda-1=0$. For selfadjoint matrices $M$, the corresponding eigenvectors are orthogonal, and thus when the vector $\mathbf{x}$ ranges over the unit ball (in this case the unit circle), $\mathbf{x}^TM\mathbf{x}$ ranges from the smallest eigenvalue to the largest, in this case, from $(3-\sqrt{13})/2$ to $(3+\sqrt{13})/2$.
On
Making $y = \lambda x$ and substituting we have
$$ \mbox{Variation for}\ \ \ x^2(3-2\lambda)\ \ \ \mbox{s. t. }\ \ x^2(1+\lambda^2) = 1 $$
or variation for
$$ f(\lambda) = \frac{3-2\lambda}{1+\lambda^2} $$
now determining the stationary points with $f'(\lambda) = 0$ obtaining
$$ \lambda = \frac 12\left(3\pm\sqrt{13}\right) $$
etc.
$3 x^2 - 2 x y$ with $y = \pm \sqrt{1-x^2}$ is $3 x^2 \pm 2 x \sqrt{1 - x^2}:$
Take derivative w.r.t. $x$, set it to zero and solve.
Note that the two functions are reflections about $x=0$, so their ranges are the same.