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I'm trying to find the critical points for $f(x, y) = 5x^2-2y^2+10$ on the circle $x^2+y^2 = 1$.

I understand that when I substitute $x = \cos(t)$ and $y = \sin(t)$, I get all four critical points: $(0, 1), (0, -1), (1, 0)$, and $(-1, 0)$. However, if I solve for $x^2$ in the constraint, I get $x^2 = 1-y^2$, and when I plug this into the function, I get $f(y) = -7y^2 + 15$. $f'(y) = -14y$, so $f'(y) = 0$ when $y = 0$.

However, this only provides $2$ of the $4$ critical points, $(0, 1)$ and $(0, -1)$. You only get the other two critical points (when $y = 1$) when you do the similar substitution for $y^2 = 1-x^2$.

Why is this? It's usually sufficient to "merge" the equations once and get the derivative to find all critical points. Why is it necessary to do so with both of them in this case?

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    $\begingroup$ On a finite interval the critical points might be realized at an endpoint. as your $f(y)$ is defined on $[-1,1]$ you need to check those two endpoints in addition to solving $f'(y)=0$. $\endgroup$ – lulu Feb 27 '18 at 22:18
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    $\begingroup$ Possibly worth remarking: Lagrange multipliers works well here, certainly avoids technical issues of the sort you encountered. $\endgroup$ – lulu Feb 27 '18 at 22:19
  • $\begingroup$ Oh yeaahhhhh totally forgot about that for one-variable functions. Thank you! $\endgroup$ – user124384 Feb 27 '18 at 22:20
  • $\begingroup$ Or actually, while I knew extrema were on the boundaries by virtue of just being the smallest or largest value, I didn't realize boundaries were also candidates for critical points and couldn't be discovered simply via derivatives like the points between the bounds. I'm still not completely sure why that would be. $\endgroup$ – user124384 Feb 27 '18 at 22:25
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    $\begingroup$ You need to distinguish between critical points—those at which the derivatives vanish—and extrema. Example: The extrema of $f(x)=x$ on any closed interval lie at its end points, but its derivative vanishes nowhere. $\endgroup$ – amd Feb 27 '18 at 22:41

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