find-the-stationary-points-of-the-function-U-x-2-y-2-subjects-to-the-constraint-x-2-y-2-2x-2y-1-0-

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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