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I'm trying to understand how to find the minimum & maximum values of this function:

$$ f(x,y) = xy-y^2 $$

In the following range D:

$$ D = \{(x,y) \in R^2 : 0 \leq x \leq 1, |y| \leq x^2 \} $$

Obviously I tried to use Lagranage multipliers, but I was a little confused about the absolute value. Should I divide it to two different equastion systems, one for positive y and one for negative?

Thanks in advance.

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observe that $$f_x=y$$ and $$f_y=x-2y$$ thus we get the solution $$x=0,y=0$$ from the system $$f_x=0$$ and $$f_y=0$$

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  • $\begingroup$ Hey, Lagrange is not required here? it seems too easy the way you showed it :) $\endgroup$ – superuser123 Jul 14 '16 at 13:17

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