Multivariable 1st Derivative Test for Critical Points
Matthew Harrington 
I am having trouble with understanding the solving method of part 2.
What is the easiest way to solve for x,y given the 2 functions from the first derivative test?
$\endgroup$ 21 Answer
$\begingroup$We have $$4x \, (1-x^2 - y^2) = 0 \tag{1}$$ $$4y \, (1-x^2 - y^2) = 0 \tag{2}$$
A good strategy might be to solve each of these equations individually and see where they match up.
It would be convenient if the $4x$ and $4y$ weren't in the equations. If they weren't there, we would be nearly finished because the system of equations would be two copies of same equation!
But the only way that could be the case is if we divided each equation by $4x$ and $4y$, respectively. But if we do that, then we have to assume that $x \not = 0$ and $y \not = 0$ (since we can't divide by $0$). So to make our solution(s) complete, we need to check whether or not $(x,y) = (0,0)$ works before we continue with this plan. Checking this ordered pair reveals that $(0,0)$ indeed satisfies the system, so we keep that as one of our solution(s).
Now we divide by $4x$ and $4y$ and we assume that $x \not = 0$ and $y \not = 0$ to get
$$1-x^2 - y^2 = 0 \tag{3}$$ $$1-x^2 - y^2 = 0 \tag{3}$$
Rearrange to get $$x^2 + y^2 = 1 \tag{4}$$
And obviously, when this is satisfied, our system of equation (sic, sort of) is satisfied.
Which means in total, our solutions are
$$x = 0, \, y = 0$$ $$x^2 + y^2 = 1$$
$\endgroup$ 4
