Question

If ${\cos }^{-1}\left(\frac{x}{2}\right)+{\cos }^{-1}\left(\frac{y}{3}\right)=\theta$, then maximum value of $9x^2-12xy\cos \theta +4y^2$ is

• A. $18$
• B. $30$
• C. $24$
• D. $36$

Answer 1

The correct option is D $36$

${\cos }^{-1}\left(\frac{x}{2}\right)+{\cos }^{-1}\left(\frac{y}{3}\right)=\theta$
$\Rightarrow {\cos }^{-1}(\frac{xy}{6}-\sqrt{(1-(\frac{x}{2}{)}^2).(1-\left(\frac{y}{3}{)}^2\right)})=\theta$
$\Rightarrow (\frac{xy}{6}-(\sqrt{(1-(\frac{x}{2}{)}^2).(1-\left(\frac{y}{3}{)}^2\right)}=\cos \theta$
$\Rightarrow (\frac{xy}{6}-\cos \theta )=\sqrt{(1-(\frac{x}{2}{)}^2).(1-\left(\frac{y}{3}{)}^2\right)}$
Squaring both side
$\Rightarrow \frac{x^2{y}^2}{36}-xy\cos \frac{\theta }{3}+{\cos }^2\theta =(1-(\frac{x}{2}{)}^2).(1-\left(\frac{y}{3}{)}^2\right)=1-\frac{x^2}{4}-\frac{y^2}{9}+\frac{x^2{y}^2}{36}$
$\Rightarrow 9x^2-12xy\cos \theta +4y^2=36(1-{\cos }^2\theta )$
Hence maximum value of $\Rightarrow 9x^2-12xy\cos \theta +4y^2$ is $36$.