I am trying to find the max and min x and y values of an ellipse. The ellipse is given by $$x^2-xy+y^2=3$$
So I need to find the max points, so I need the derivative of the function.
I got $$\dfrac{dy}{dx} = \dfrac{(y-2x)}{(2y-x)}$$ So I then thought that I should get the x value and the y value so I got $x=-(y/2)$ and $y=-2x$
But now I'm not sure what to do?
First determine when $\dfrac {dy}{dx} = 0$, in terms of $x$ and in terms of $y$. In terms of $y$, this is when $y = 2x$, and in terms of $x$, this is when $x = \frac y2$. (You simply have sign-errors in your conclusions).
Next, substitute $y = 2x$ into the equation of the ellipse, and solve the resulting quadratic equation to find it's zeros. Then determine which, if either, gives a maximum, which is a minimum.
$$x^2-xy+y^2=3 \iff x^2 - x(2x) + (2x)^2 - 3 = 0$$ $$ \iff 3x^2 - 3 = 3(x^2 - 1)= 0$$
$$x = \pm 1 $$
Now, find the corresponding $y$ values, using the equation of the parabola $$x = 1 \implies y^2 - y - 2 = (y - 2)(y + 1) = 0\implies y = 2 \text{ or } y = -1$$ $$x = -1 \implies y^2 + y - 2 = (y + 2)(y - 1) = 0 \implies y = -2 \text{ or } y = 1$$
So it looks like the point at which $(x, y) = (1, 2)$ is a maximum, and the point $(x, y) = (-1, -2)$ is a minimum. Double check. This is consistent with the determination that the first derivative is zero when $y = 2x$: $x = 1 \implies y = 2$, and $x = -1 \implies y = -2$.