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 .
 A: If $64= (3y^2)^2 + 4x^4$, then $3y^2 =\sqrt{ 64 - 4x^4}$. Plugging that to $x^2+y^2$ we get:
$$x^2+\frac{\sqrt{64 - 4x^4}}{3}.$$
Now you have to maximalize a function of one variable. Remember that if $64 = 4x^4+9y^4$, then (setting $y = 0$) we obtain $16 \ge x^4$, therefore $|x| \le 2$.
A: $x^2=s,\ y^2=t \geq 0$ then $$ (2s)^2+ (3t)^2=8^2 \Rightarrow
2s=4\cos\ \theta,\ 3t=4\sin\ \theta,\ 0\leq\theta \leq
\frac{\pi}{2} $$
Then $$ s+t=2\cos\ \theta + \frac{4}{3}\sin\ \theta=\sqrt{2^2+
\bigg(\frac{4}{3}\bigg)^2} \sin\ (\theta+\alpha),\ 0< \alpha <
\frac{\pi}{2} $$
Let $\theta =\frac{\pi}{2}-\alpha$ 
A: Without using Lagrange multipliers
It's equivalent to find $x$ that maximizes
$$f(x)=x^2+\frac23\sqrt{16-x^4}$$
You have
$$f'(x)=2x-\frac13\frac{4x^3}{\sqrt{16-x^4}}$$
Hence $f'(x)=0$ iff $x=0$ or
$$\frac{2x^2}{\sqrt{16-x^4}}=3$$
$$2x^2=3\sqrt{16-x^4}$$
$$4x^4=9(16-x^4)$$
$$13x^4=12^2$$
$$x=\pm2\frac{\sqrt{3}}{\sqrt[4]{13}}$$
Let $x_0=2\dfrac{\sqrt{3}}{\sqrt[4]{13}}$. Since $x$ appears only with even powers, sign is not important. And you have
$$f(0)=\frac83$$
$$f(x_0)=\frac{12}{\sqrt{13}}+\frac{2}{3}\sqrt{16-\frac{16\times 9}{13}}$$
$$=\frac{12}{\sqrt{13}}+\frac{2}{3}\sqrt{\frac{64}{13}}=\frac{12+\frac{16}{3}}{\sqrt{13}}=\frac{4\times13}{3\sqrt{13}}=\frac{4}{3}\sqrt{13}$$
And this is larger than $f(0)$ since
$$f(0)^2-f(x_0)^2=\frac{16\times4}{9}-\frac{16\times13\times3}{9}$$
Thus the max is found for $x=\pm x_0$ and
$$y^4=\frac{1}{9}\left(64-4x_0^4\right)=\frac{1}{9}\left(64-4\frac{16\times 9}{13}\right)=\frac{64}{9}\left(1-\frac{9}{13}\right)=\frac{64\times4}{9\times13}$$
Hence
$$y=\pm \frac{4}{\sqrt{3}\sqrt[4]{13}}$$
A: Since you tagged Lagrange Multipliers, I will use that method.
Allow that $F(x,y) = 4x^4+9y^4-64$ and $G(x,y)=x^2+y^2-C$, where $C$ is treated as a constant and is the maximum you wish to find.
We allow that $\nabla F(x,y) = \lambda \nabla G(x,y)$ and find two new equations:
$16x^3 = \lambda 2x$ and
$36y^3 = \lambda 2y$
$\implies x^4 = \frac{\lambda ^2}{64}$ and $y^4 = \frac{\lambda ^2}{324}$.
Plugging those into $F(x,y)$ we find that:
$$\lambda = 26.6256...$$
We now go back to our equations after applying the gradient and solve for $x_{max}$ and $y_{max}$. We get:
$x_{max}=1.8245...$ and $y_{max}=1.21622...$
Plugging this into $x^2+y^2$ we obtain the maximum, $4.80727$.
A: 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 $(x,y) = (2,0)$, $$x^2 + y^2 = 2^2 + 0^2 = 4 < \frac{4}{3}\sqrt{13} \ \text{ as } 16 < \frac{16}{9}\cdot 13$$
