lagrange multipliermultivariable-calculus
I know I have to use the method of Lagrange multipliers so I want to find $\lambda$ in
$$\langle 2xy,x^2 \rangle=\lambda \langle2x,4y \rangle$$
This gives me $y =\lambda$ but my multivariable calc is very rusty. Do I use $\lambda$ to find $x,y$ to plug into $x^2y$?
If you note that $x\neq 0$ and $y\neq 0$ you can divide the second equation by the third and get:
$$\frac{2ye^{2xy}}{2xe^{2xy}} = \frac{2\lambda x}{2\lambda y} \ ,$$ which after simplification says that $x^2=y^2$. Taking the first equation, you can then write $$x^2 + y^2 = 2x^2 = 16$$ This gives $x=\pm 2\sqrt{2}$ and therefore four possible solutions $(x,y)=(\pm 2\sqrt{2},\pm 2\sqrt{2})$, where all sign combinations are possible.
Lagrange multipliers do work for your problem, but that involves solving three simultaneous non-linear equations in three unknowns. There is another way that uses only one variable: parameterization.
Get a parameterization that describes the given curve in terms of only one variable. In the case of your ellipse you can use
$$x=\cos t, \quad y=\frac 17\sin t, \quad 0\le t\le 2\pi$$
Now you want to want to find the extrema of
$$\hat f(t)=4x+y=4\cos t+\frac 17\sin t$$
There are several ways to find the extrema of that, using calculus or just trigonometry. Here is a calculus way:
$$\hat f'(t)=-4\sin t+\frac 17\cos t=0$$ $$28\sin t=\cos t$$ $$\tan t=\frac 1{28}$$ $$\cos t=\sqrt{\frac 1{\tan^2 t+1}}=\pm\frac{28}{\sqrt{785}}$$ $$\sin t=\sqrt{1-\cos^2 t}=\pm\frac{1}{\sqrt{785}}$$ $$x=\cos t=\pm\frac{28}{\sqrt{785}}$$ $$y=\frac 17\sin t=\pm\frac{1}{7\sqrt{785}}$$
where $x$ and $y$ have the same sign. This means the maximum of $f$ is
$$4x+y=4\frac{28}{\sqrt{785}}+\frac{1}{7\sqrt{785}}=\frac{785}{7\sqrt{785}}=\frac{\sqrt{785}}{7}\approx 4.00255$$
and the minimum is the negative of that. This graph confirms the maximum.
Best Answer
As I mentioned in the comments, if this isn't a homework problem for which you'd lose points for not using Lagrange multipliers, it would be better just to use substitution to change it to a single-variable calculus problem.
In particular, note $x$ only appears as $x^2$ and we may solve $x^2 + 2y^2=6$ for $x^2 = 6-2y^2$. We may now maximize $$x^2y = (6-2y^2)y$$ subject to the condition $6-2y^2 = x^2 \geq 0$. That is, maximize $f(y) = 6y-2y^3$ on the interval $[-\sqrt{3}, \sqrt{3}]$.
Setting the derivative equal to zero to find the critical points gives $$0=f'(y)= 6-6y^2 \implies y=\pm 1$$ and then checking these points gives a maximum of $f(1) = 4$.