Given $ 4x^2 +2y^2 =1 $ then find the maximum value of $ 4x+2y $
My thinking: I have taken $ x= \frac{sin{\theta}}{2} $ and $y=\frac{cos{\theta}}{\sqrt{2}}$ . Which satisfies the equation. After substitution I've got $4x+2y= 2\times sin{\theta} +\sqrt{2} \times cos{\theta} $ which has the critical point at $\theta=tan^-1{{\sqrt{2}}}$. So, which gives the maximum value is $\sqrt{6}$
Help me with the correct procedure. Thanks in advance.
Let $2x=\sin t,\sqrt2y=\cos t$
$$4x+2y=2\sin t+\sqrt2\cos t=\sqrt{2^2+2}\sin\left(t+\arctan\dfrac{\sqrt2}2\right)\le\sqrt6$$
Another way :
Let $$4x+2y=k\iff x=?$$
Replace the value of $x$ in $$4x^2+2y^2=1$$ to form a Quadratic Equation in $z$
As $z$ is real, the discriminant must be $\ge0$
You could also use Lagrangian multipliers. With a Lagranagian $L=4x+2y-\lambda(4x^2+2y^2-1)$, we simultaneously solve $0=\frac{\partial L}{\partial x}=4-8x\lambda,\,0=\frac{\partial L}{\partial y}=2-4y\lambda$ with$$x=y=\frac{1}{2\lambda}\implies x^2=y^2=\frac16,\,4x+2y\le\sqrt{6}.$$
Good old substitution works. You have that $$4x+2y=4x+2\left(\sqrt{\frac12-2x^2}\right).$$ Differentiating then gives $$4+2\frac12\frac{-4x}{\sqrt{\frac12-2x^2}},$$ and setting this to $0$ we obtain $$x=\sqrt{\frac12-2x^2},$$ which occurs when $$x=\pm\frac{1}{\sqrt 6}.$$
Since the original function can only decrease when $x<0,$ then a potential maximum occurs at the positive root obtained above. Finally, compare with what you have at the endpoints of the domain $|x|\le 1/2$ to obtain your global maximum.
Continue by applying the C-S inequality
$$4x+2y= 2\sin{\theta} +\sqrt{2} \cos{\theta} \le \sqrt{ (2^2 + (\sqrt2)^2)(\sin^2\theta+\cos^2\theta)}= \sqrt6 $$
where the equality, or the maximum value $\sqrt6$, occurs at $\frac2{\sqrt2}=\frac{\sin\theta}{\cos\theta}$.
Let $$4x+2y=k\iff x=\frac{k-2y}{4}$$
After replacing the value of $x$ in $$4x^2+2y^2=1$$, we have a quadartic equation
As $z$ is real, the discriminant has to be $\ge0$
