Find-maximum-and-minimum-values-of-f-x-y-xy-on-the-circle-g-x-y-x-2-y-2-1-0-

Question Number 122695 by liberty last updated on 19/Nov/20

F i n d m a x i m u m a n d m i n i m u m v a l u e s

o f f (x, y) = x y o n t h e c i r c l e g (x, y) =

x^{2} + y^{2} - 1 = 0 .

Answered by john santu last updated on 19/Nov/20

B y L a g r a n g e m u l t i p l i e r

▽ f (p) = λ ▽ g (p), w h e r e P i s a n y p o i n t

(\begin{matrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{matrix}) = λ (\begin{matrix} \frac{\partial g}{\partial x} \\ \frac{\partial g}{\partial y} \end{matrix})

(\begin{matrix} y \\ x \end{matrix}) = (\begin{matrix} 2 λ x \\ 2 λ y \end{matrix}) \to {\begin{cases} λ = \frac{y}{2 x} \\ λ = \frac{x}{2 y} \end{cases}

\Rightarrow \frac{y}{2 x} = \frac{x}{2 y} \Rightarrow x^{2} = y^{2} \land x^{2} + y^{2} = 1

t h i s g i v e s f o u r s o l u t i o n s

{\begin{cases} (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}), (\frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}}) \\ (- \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}), (- \frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}}) \end{cases}

E v a l u a t i n g f a t t h e s e f o u r p o i n t s

w e g e t m a x v a l u e i s \frac{1}{2} a n d m i n v a l u e

i s - \frac{1}{2} .

Answered by mr W last updated on 19/Nov/20

P = x y

y = \frac{P}{x}

x^{2} + \frac{P^{2}}{x^{2}} - 1 = 0

2 x + \frac{2 {P P}^{'}}{x^{2}} - \frac{2 P^{2}}{x^{3}} = 0

P^{'} = 0

x - \frac{P^{2}}{x^{3}} = 0

x - \frac{x^{2} y^{2}}{x^{3}} = 0

x^{2} = y^{2} = \frac{1}{2}

\Rightarrow x = \pm \frac{1}{\sqrt{2}}, y = \pm \frac{1}{\sqrt{2}}

{(x y)}_{m a x} = \frac{1}{2}

{(x y)}_{m i n} = - \frac{1}{2}

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