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 .

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
$$y=\pm \frac{4}{\sqrt{3}\sqrt[4]{13}}$$
