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I'm trying to get the extrema of a function $x + y²$ with a constraint $2x² + y² = 1$ using Lagrange multipliers. The Lagrange function is $x + y² + \lambda (2x² + y² - 1)$. I have three partial derivatives:

  1. $$1 + 4\lambda x = 0$$
  2. $$2y + 2\lambda y = 0$$
  3. $$2x² + y² - 1 = 0$$

My approach to solving these for x and y is as follows:

  1. $$ 2\lambda = - 1 / 2x$$
  2. $$ 2\lambda = - 2y / y = -2$$ $$-1/2x = -2$$ $$x = 1/4 $$ Plugging into the constraint: $$2(1/4)² + y² = 1$$ $$1 - 2/16 = y²$$ $$y = \pm \sqrt{7/8}$$ So I get 2 points $(1/4, -\sqrt{7/8})$ and $(1/4, \sqrt{7/8})$ which are the correct maxima. If I plug the equation system into Wolfram Alpha, I get two more solutions, which I understand are the minima. How do I get these two other solutions? I'm probably missing some edge case, but I'm not sure how else to solve the equation system.
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    $\begingroup$ Did you copy some wrong numbers? Your derivatives do not match with your original problem. $\endgroup$ – KittyL Jun 4 '15 at 13:37
  • $\begingroup$ Right, sorry, I mixed up the pages while copying. Fixed. $\endgroup$ – 1nterference Jun 4 '15 at 13:45
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When you are doing $$2\lambda = - 2y / y = -2$$ You lost the potential solution $y=0$. That would give you another answer.

When doing Lagrange multiplier problems, be careful about all the cases, especially when multiplying or dividing by a variable.

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