Question

Let $x,y,z,t$ be real numbers $x^2+y^2=9,z^2+t^2=4$, and $xt-yz=6$. Then the greatest value of $P=xz$ is

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

The correct option is B $3$

Let $x,y,z,t,\theta ,\varphi \in R$ with

$x^2+y^2=9;z^2+t^2=4;xt-yz=6$

The following substitutions are valid;

$x=3\cos \theta ,y=3\sin \theta ,z=2\cos \varphi ,t=2\sin \varphi$

$xt-yz=6$

$\Rightarrow \left(3\cos \theta \right)\left(2\sin \varphi \right)-\left(3\sin \theta \right)\left(2\cos \varphi \right)=6$

$\Rightarrow 6\sin \varphi \cos \theta -6\cos \varphi \sin \theta =6$

$\Rightarrow \sin (\varphi -\theta )=1$

$\Rightarrow \varphi -\theta =2n\pi +\frac{\pi }{2}$ or $(2m+1)\pi -\frac{\pi }{2}m,n\in z$

$xz=\left(3\cos \theta \right)\left(2\cos \varphi \right)$

$=6\cos \theta \cos \varphi$

$=6\cos \theta \cos (2n\pi +\frac{\pi }{2}+\theta )$

$=-6\cos \theta \sin \theta$

$=-3\sin 2\theta$

(or)

$xz=6\cos \theta \cos \left(\right(2m+1)\pi -\frac{\pi }{2}+\theta )$

$=-6\cos \theta \sin \theta$

$=-3\sin 2\theta$

$(xz{)}_{max}=3$ and $(xz{)}_{min}=-3$