Greatest value of $xz$ in expression 
Let $x,y,z,t\in\mathbb{R}$ and $x^2+y^2=9,z^2+t^2=4$ and $xt-yz=6$. Then greatest value of $xz$ is 

What i try
I am trying to solve without trigonometric substution
$$(xt-yz)^2+(xz+yt)^2=(x^2+y^2)(z^2+t^2)$$
$6^2+(xz+yt)^2=4\cdot 9\Longrightarrow (xz+yt)^2=0$
$$xz=-yt$$
How do i solve it Help me please
 A: Notice that:
$$4(x^2+y^2) + 9(t^2 + z^2) = 12(xt - yz)$$
and this is equivalent with
$$(2x - 3t)^2+ (2y + 3z)^2 = 0$$
Thus $z=-\frac{2}{3}y$, and we want to maximize $-\frac{2}{3}xy$. This is simple, given that $x^2+y^2=9$ because:
$$0\leq (x+y)^2\Rightarrow -2xy \leq x^2+y^2=9$$
Therefore:
$$zx=-\frac{2}{3}xy\leq 3$$
with equality when $(x,z)=\left(\pm \frac{3}{\sqrt{2}},\pm \sqrt{2}\right)$.
A: A bit late answer but I think worth mentioning it.
You can find the searched for maximum exploiting the relation you have already found:


*

*$xz=-yt$
You can do this, for example, by using basic facts about complex numbers writing


*

*$u = x+iy$ and $v = t+iz \Rightarrow |u|=3, |v|=2$

*$\Rightarrow 6= xt-yz = \Re(uv) =|uv|=|u||v|=6$

*$\Rightarrow \Im(uv)=xz+yt  =0$ (The intermediate result you have already found.)


Now, we can exploit this by considering


*

*$\Im(\bar u v) = xz-ty \stackrel{xz=-yt}{=}2xz \leq |\bar u v|=6$

*$\Rightarrow \boxed{xz \leq 3}$
Addendum concerning the maximum:
You are either fully aware that the maximum must be reached. Otherwise, it is easy to indicate specific $u,v$ and, hence, $x,z$ for which the maximum is reached:


*

*$u=3, v=2$ obviously satisfy the initially given condition.

*$\Rightarrow u_{\phi}=3e^{i\phi}, v_{\phi}=2e^{-i\phi}$ satisfy the initially given condition, as well.

*$\Rightarrow \Im(\bar u_{\phi}v_{\phi}) = 6\Im\left(e^{-2i\phi}\right)=6$ for $\phi = -\frac{\pi}{4}$

*$\Rightarrow$ Maximum is reached for 
$$u= 3e^{-i\frac{\pi}{4}} = \underbrace{\frac{3\sqrt{2}}{2}}_{=x}-i\frac{3\sqrt{2}}{2}, v=2e^{i\frac{\pi}{4}} = \sqrt{2}+i\underbrace{\sqrt{2}}_{=z}$$
