Find the minimum of the value $3x^2-2xy$ if $\frac{x^2}{4}-y^2=1$ Let $x,y\in R$,such
$$\dfrac{x^2}{4}-y^2=1$$
find the minium of the 
$$3x^2-2xy$$
I think $x=2\sec{t},y=\tan{t}$,then
$$3x^2-2xy=12(\sec{t})^2-4\sec{t}\tan{t}$$
 A: hint 
using the fact that
$$\cosh^2 (t)-\sinh^2 (t)=1$$
we will put
$$x=2\cosh (t) $$ and
$$y=\sinh (t) $$
then
$$f (t)=3x^2-2xy$$
$$=4\cosh (t)(3\cosh (t)-\sinh (t)) $$
$$=6 (1+\cosh (2t))-2\sinh (2t) $$
You can solve $f'(t)=0$.
A: Let $x=2(u+v)$ and $y = u-v$. Then $\frac{x^2}{4}-y^2 = 1 \Rightarrow uv = \frac{1}{4}$
Now $3x^2 - 2xy = 8u^2+16v^2+24uv = 8\left(u^2+\frac{1}{8u^2} \right)+6 \ge 4\sqrt 2+6$ by AM-GM Inequality
A: Let $3x^2-2xy=k$.
Since $x^2=4y^2+4$, we obtain:
$$k=3x^2-2xy\geq3x^2-(x^2+y^2)=2(4+4y^2)-y^2=8+7y^2\geq8.$$
Hence, 
$$3x^2-2xy=k\left(\frac{x^2}{4}-1\right)$$ or
$$\left(3-\frac{k}{4}\right)x^2-2xy+ky^2=0,$$
which gives 
$$1-k\left(3-\frac{k}{4}\right)\geq0$$ or
$$\frac{k^2}{4}-3k+1\geq0,$$ which is
$k\geq2(3+2\sqrt2)$ or $k\leq2(3-2\sqrt2)$ and since $k\geq8$, we obtain:
$$k\geq2(3+2\sqrt2).$$
The equality occurs for $k=2(3+2\sqrt{2})$, $\frac{x^4}{4}-y^2=1$ and $\frac{x}{y}=\frac{1}{3-\frac{2(3+2\sqrt2)}{4}}=2(3+2\sqrt2),$
which says that the equality occurs.
Thus, we got the answer:
$$\min_{\frac{x^2}{4}-y^2=1}(3x^2-2xy)=2(3+2\sqrt2).$$
Done!
A: HINT: you can solve your condition $$\frac{x^2}{4}-1=y^2$$ for $y$ $$y=\pm\sqrt{\frac{x^2}{4}-1}$$ and plug this in your objective function $$f(x)=3x^2-2x\left(\pm\sqrt{\frac{x^2}{4}-1}\right)$$ and you will get a Problem in only one real varible. 
A: NOTE: I've just noticed @Salahamam_Fatima's answer...
Let $f(x, y)$ be the function you want to minimize. Observe that $f(x, y) = f(-x, -y)$. So let's restrict ourselves to $x \geq 0$. The condition $\displaystyle\frac{x^2}{4} - y^2 = 1$ gives you $x^2 \geq 4$ and hence $x \geq 2$. Solving a degree $2$ equation in $e^t$, we get that there exists $t \in \mathbb{R}$ such that $x = e^t + e^{-t}$. By possibly interchanging $t$ with $-t$, we get that $y = \displaystyle\frac{1}{2}(e^t - e^{-t})$. So now you need to minimize a function of just one variable. Namely, let $g(t) = 3(e^t + e^{-t})^2 - 2(e^t + e^{-t})\displaystyle\frac{1}{2}(e^t - e^{-t}) = e^{2t} + 4e^{-2t} + 6$. This can be simplified even more by replacing $e^{t}$ with $u$. We know have to minimize $h(u) = u^2 + \displaystyle\frac{4}{u^2} + 6 = (u - \displaystyle\frac{2}{u})^2 + 10 \geq 10$. Hope I didn't mess up any calculations. 
A: This sure looks like a Lagrange multiplier problem to me.  Let $f(x,y) = 3x^2-2xy$ and $g(x,y) = \frac{x^2}{4}-y^2.$  Then $\nabla f = (6x-2y,-2x)$ and $\nabla g = (\frac{x}{2}, -2y)$.  We get the system of 3 equations in three variables:
$$6x-2y = \lambda \frac{x}{2}$$
$$-2x = \lambda (-2y)$$
$$\frac{x^2}{4}-y^2 = 1$$
After some algebraic juggling, there are $8$ solutions:
$$x= \pm\frac{1}{3}\sqrt{18\pm 6\sqrt{10}}, y=  \mp \frac{1}{6}\sqrt{18\pm 6\sqrt{10}}(-3\pm\sqrt{10}),$$
where there are $4$ choices for $x$ and for each one, two corresponding choices for $y$.  
If all $8$ points are plugged into $f$, there are $4$ results: $$ \pm 2\sqrt{10}\pm\frac{20}{3},$$
and the max and min are obvious.
A: Rearranging B. Goddard's system of equations:
$$\begin{align}12x-4y-\lambda x=0\end{align}\tag1$$
$$\begin{align}x=\lambda y\end{align}\tag2$$
$$\begin{align}x^2-4y^2=4\end{align}\tag3$$
Inserting $(2)$ into $(1)$ yields
$$\begin{align}12\lambda y - 4y - \lambda^2y=0\end{align}$$
$\Leftrightarrow$
$$\begin{align}(\lambda^2-12\lambda + 4)y=0\end{align}.$$
The two solutions to the quadratic equation are 
$$\begin{align}\lambda_1=-4\sqrt 2 + 6\end{align}$$
and
$$\begin{align}\lambda_2=+4\sqrt 2 + 6\end{align}.$$
Inserting $(2)$ into $(3)$ gives us
$$\begin{align}(\lambda^2-4)y^2=4\end{align}.$$
For $\lambda_1$ with
$$\begin{align}\lambda_1^2=-48\sqrt 2+68\end{align}$$
and
$$\begin{align}\lambda_1^2-4=16(-3\sqrt 2+4)\end{align}$$
so that
$$\begin{align}\dfrac{4}{\lambda_1^2-4}=\dfrac{1}{4(-3\sqrt 2+4)}=\dfrac{3\sqrt 2+4}{4\cdot-2}\end{align}$$
it follows that
$$\begin{align}-8y^2=+3\sqrt{2}+4\end{align}.$$
Unfortunately this is not solvable over the reals.
Likewise for $\lambda_2$ we have
$$\begin{align}-8y^2=-3\sqrt{2}+4\end{align}$$
which is solvable over the reals since $3\sqrt{2}\geq 4$.
The minimum is 
$$\begin{align}3x^2-2xy=4\sqrt 2 + 6\end{align}.$$
