This is kind of stupid since I already have the answers, but I feel like I overlooked an important aspect of Lagrange Multipliers to misuse it here.

Given $f = x^2 + xy + x - y$ and the constraint is the region formed by $y = 1 - x$ and the coordinate axis.

Now I realize I havce to test it alone the boundary, but could someone explain to me what I overlooked when I want to use Lagrange Multipliers?

I was thinking of apply Lagrange Multipliers to each constraint. So for instance on the line $y + x = g(x,y) = 1$ I would do

$\nabla f = \lambda \nabla g$

$g = 1$

After doing so I got a contradiction where x + y = 1 and x + y = 0


You didn't get a contradiction. You got a system of equations with no solution. Therefore, on the line $x+y=1$, there are no critical points.

Don't forget that you were trying to optimize not on the entire line, but on the subset of the line $x+y=1$ with $0 \leq x \leq 1$, so you need to check the boundary points too. As there are no critical points, the optima of your function constrained to this line segment occur on the boundary of this line segment.

  • $\begingroup$ I got f(0,1) = -1 and f(1,0) = 2 testing along the boundary points. I thought there's a theorem that says if I bound the region then I am bound to get extremas $\endgroup$ – Hawk Jun 3 '12 at 4:42
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    $\begingroup$ A continuous function on a closed, bounded region achieves a maximum and a minimum, yes. But, where? Could be at those boundary points you tested. No theorem says there's a critical point anywhere. $\endgroup$ – Gerry Myerson Jun 3 '12 at 4:49

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