Maximum value of $ x^2 + y^2 $ given $4 x^4 + 9 y^4 = 64$ It is given that $4 x^4 + 9 y^4 = 64$.
Then what will be the maximum value of $x^2 + y^2$?
I have done it using the sides of a right-angled triangle be $2x , 3y $ and hypotenuse as 8 .

One more stab at the problem. Instead of working with $x$, let's work with $u = x^2$ as it makes the algebra slightly easier. 
As $4x^4 + 9y^4 = 64$, we have $y^2 = \frac{1}{3}\sqrt{64-4u^2}$, taking only the positive square roots as $y^2 \geq 0$. Hence want to maximize
$$f(u) = x^2 + y^2 = u + y^2 = u + \frac{1}{3}\sqrt{64-4u^2}$$
Setting the first derivative equal to zero,
$$0 = \frac{df}{du} = 1 - \frac{4}{3}u \frac{1}{\sqrt{64-4u^2}}$$
$$\Longrightarrow  \  64 - 4u^2 = \frac{16}{9}u^2 \hspace{30 mm}$$
$$\Longrightarrow \ \ u^2 = \frac{9}{52}\cdot 64 \ \ \text{ i.e., } u = \frac{12}{\sqrt{13}} \hspace{10 mm}$$
(We take only the positive square root in the last step as $u = x^2 \geq 0$.) 
Using the earlier expression for $y^2$, we find for $u = x^2 = \frac{12}{\sqrt{13}}$ that
$$y^2 = \frac{1}{3}\sqrt{64 - 4\cdot \frac{144}{13}} = \frac{16}{3\sqrt{13}}$$
Hence
$$x^2 + y^2 = \frac{12}{\sqrt{13}} + \frac{16}{3\sqrt{13}} = \frac{52}{3\sqrt{13}} = \frac{4}{3}\sqrt{13}$$
This is a maximum as there are other points satisfying the constraint for which $x^2 + y^2 < \frac{4}{3}\sqrt{13}$. For example, for $(x,y) = (2,0)$, $$x^2 + y^2 = 2^2 + 0^2 = 4 < \frac{4}{3}\sqrt{13} \ \text{ as } 16 < \frac{16}{9}\cdot 13$$
