Question

If $4x^2+y^2=1,$ then the maximum value of $12x^2-3y^2+16xy$ is

Answer 1

$4x^2+y^2=1,$
$\Rightarrow \frac{x^2}{{\left(\frac{1}{2}\right)}^2}+\frac{y^2}{1^2}=1$

Which is an equation of ellipse, hence any point on it can be written as
$x=\frac{1}{2}\cos \theta ,y=\sin \theta$

$\therefore 12x^2-3y^2+16xy=12{\left(\frac{\cos \theta }{2}\right)}^2-3(\sin \theta {)}^2+16\left(\frac{\cos \theta }{2}\right)(\sin \theta )=3{\cos }^2\theta -3{\sin }^2\theta +8\sin \theta \cos \theta =3\cos 2\theta +4\sin 2\theta$

As we know that,
$-5\le 3\cos 2\theta +4\sin 2\theta \le 5$
So, the maximum value is $5$