Maximum value of $x^2+y^2$

Question

Question

Find the maximum value of $x^2+y^2$ if $4x^4+9y^4=64$

Now I really don't understand how to proceed or whether I should change my approach all together. Any help is appreciated.Thanks :)

Answer 1

WLOG, $2x^2=8\cos t,3y^2=8\sin t$ where $0\le t\le\frac\pi2$

Answer 2

$$4x^4+9y^4=64$$

$$\frac{(x^2)^2}{4^2}+\frac{(y^2)^2}{(8/3)^2}=1$$

Let $x=\pm \sqrt{4\cos(\theta)}$ and $y=\pm \sqrt{\frac{8}{3} \sin (\theta)}$. With $\theta \in [0,\frac{\pi}{2}]$.

Then,

$x^2+y^2=4\cos (\theta)+\frac{8}{3} \sin (\theta)$

$$=\langle 4, \frac{8}{3} \rangle \cdot \langle \cos (\theta), \sin (\theta) \rangle$$

$$=\sqrt{4^2+(\frac{8}{3})^2} \cos \left( \theta-\arctan(\frac{\frac{8}{3}}{4}) \right)$$

Because $\arctan(\frac{8}{12}) \in [0, \frac{\pi}{2}]$. A maximum of,

$$\sqrt{4^2+(\frac{8}{3})^2}$$

Is achievable.