Maximizing $3 x^2+2 \sqrt{2} x y$ with $x^4+y^4=1$

Question

I want to have maxizing value of $3 x^2+2 \sqrt{2} x y$ when $x^4+y^4=1$, $x>0,y>0$. How can I solve it.

Answer 1

here is one way to do this: let us parametrize the constraint $$x^4 + y^4 = 1, x > 0, y > 0 \text{ by } x = \sqrt{\cos t}, y = \sqrt{\sin t}, 0 < t < \pi/2.$$ we need to maximise $$ f = 3x^2 + 2\sqrt 2xy=3\cos t + 2\sqrt 2\sqrt{\cos t \sin t} = 3\cos t + 2\sqrt{\sin 2t}, 0 < t < \pi/2$$ taking the derivative of $f$ and seeting it to zero, you have $$f' = -3\sin t + 2\frac{\cos 2t}{\sqrt{\sin 2t}}$$

only critical number of $f$ is $t = 0.4636$ and this is a global max $4.4721$ for $f(t),\, 0 \le t \le \pi/2.$

Answer 2

I have got solution with precaculus method. thank you everyone advices.

$3 x^2+2 \sqrt{2} x y$

$\leq$ $3 x^2+\left(x^2+2 y^2\right)$

$=$ $4 x^2+2 y^2$ (when equal is $x=\sqrt{2} y$ with arithmetic geometric mean inequality )

$=$ $2\cdot \left(2 x^2\right)+1\cdot \left(2 y^2\right)$

$\leq$ $\sqrt{\left(2^2+1^2\right) \left(\left(2 x^2\right)^2+\left(2 y^2\right)^2\right)}$

$=$ $\sqrt{\left(2^2+1^2\right) \left(4 x^4+4 y^4\right)}$ (when equal is $x=\sqrt{2} y$ with inequality of de Cauchy-Schwarz)

$=$ $\sqrt{5\cdot 4}$

$=$ $2 \sqrt{5}$

And I also can got the the solution with comment of @Théophile .