find the stationary points of the function U=x^2 +y^2 subjects to the constraint x^2 +y^2 +2x−2y+1=0


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Question Number 115922 by mathdave last updated on 29/Sep/20

$f i n d t h e s t a t i o n a r y p o i n t s o f t h e$ $f u n c t i o n U = x^{2} + y^{2} s u b j e c t s t o$ $t h e c o n s t r a i n t$ $x^{2} + y^{2} + 2 x - 2 y + 1 = 0$
	Answered by bemath last updated on 29/Sep/20

	$f (x, y, λ) = x^{2} + y^{2} + λ (x^{2} + y^{2} + 2 x - 2 y + 1)$ $\frac{\partial f}{\partial x} = 2 x + λ (2 x + 2) = 0 \to λ = \frac{- 2 x}{2 x + 2}$ $\frac{\partial f}{\partial y} = 2 y + λ (2 y - 2) = 0 \to λ = \frac{2 y}{2 - 2 y}$ $\frac{\partial f}{\partial λ} = x^{2} + y^{2} + 2 x - 2 y + 1 = 0$ $\Leftrightarrow \frac{x}{2 x + 2} = \frac{y}{2 y - 2} \Rightarrow 2 x y - 2 x = 2 x y + 2 y$ $y = - x \Rightarrow x^{2} + x^{2} + 2 x + 2 x + 1 = 0$ $2 x^{2} + 4 x + 1 = 0; x^{2} + 2 x + \frac{1}{2} = 0$ ${(x + 1)}^{2} - \frac{1}{2} = 0 \Rightarrow x = \pm \frac{1}{\sqrt{2}} - 1$ $a n d y = \mp \frac{1}{\sqrt{2}} + 1$
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