If $4{\mathrm{x}}^2+{\mathrm{y}}^2=1$ then the maximum value of $12{\mathrm{x}}^2-3{\mathrm{y}}^2+16\mathrm{xy}\text{ is }(\mathrm{x},\mathrm{y}\in \mathrm{R})$

 Let any point on $4{\mathrm{x}}^2+{\mathrm{y}}^2=1 \mathrm{is} \left(\frac{\cos {\displaystyle }\mathrm{\theta }}{2},\sin \mathrm{\theta }\right)$
$$\begin{gathered} 12{\mathrm{x}}^2-3{\mathrm{y}}^2+16\mathrm{xy}=12\frac{{\cos }^2\mathrm{\theta }}{4}-3{\sin }^2\mathrm{\theta }+16\frac{\cos \mathrm{\theta }}{2}\sin \mathrm{\theta } \\ \begin{array}{l}=3\left({\cos }^2\mathrm{\theta }-{\sin }^2\mathrm{\theta }\right)+8\cos \mathrm{\theta sin}\mathrm{\theta }\\ =3\cos 2\mathrm{\theta }+4\sin 2\mathrm{\theta }\end{array} \\ \text{ Maximum value }=\sqrt{3^2+4^2}=5 \end{gathered}$$