Determine-the-maximum-and-minimum-of-the-function-f-x-y-x-4-4x-2-y-2-2x-2-2y-2-1-Thank-you-

Question Number 203526 by Mastermind last updated on 21/Jan/24

Determine the maximum and minimum of

the function :

f (x, y) = x^{4} + {4 x}^{2} y^{2} - {2 x}^{2} + {2 y}^{2} - 1

Thank you

Answered by aleks041103 last updated on 21/Jan/24

l e t u s s e e f o r w h i c h c \in R, \exists (x, y) : f (x, y) = c

x^{2} = a, y^{2} = b

\Rightarrow f = a^{2} + 4 a b - 2 a + 2 b - 1 = c

(a^{2} - 2 a - 1) + 2 b (2 a + 1) = c

\Rightarrow b = \frac{- a^{2} + 2 a + c + 1}{2 a + 1} = y^{2} ⩾ 0

a l s o a = x^{2} ⩾ 0

- a^{2} + 2 a + c + 1 = 0 \Rightarrow a_{1, 2} = \frac{- 2 \pm \sqrt{8 + 4 c}}{- 2} = 1 \pm \sqrt{2 + c}

2 a + 1 = 0 \Rightarrow a_{3} = - 1 / 2

\Rightarrow a ⩾ 0, 2 a + 1 > 0

1 s t c a s e : c < - 2 \Rightarrow - a^{2} + 2 a + 1 + c < 0

\Rightarrow b < 0 \to c o n t r a d i c t i o n

2 n d c a s e : c = - 2 \Rightarrow - a^{2} + 2 a + 1 + c ⩽ 0

w h e r e f o r a = 1 a n d b = 0

a n d f o r a > 0, b < 0 \to c o n t r a d i c t

3 r d c a s e : - 1 ⩾ c > - 2

\Rightarrow b ⩾ 0 f o r 1 - \sqrt{2 + c} ⩽ a ⩽ 1 + \sqrt{2 + c}

4 t h c a s e : c > - 1

\Rightarrow b ⩾ 0 f o r 0 ⩽ a ⩽ 1 + \sqrt{2 + c}

\Rightarrow f o r a l l c ⩾ - 2, \exists (x, y) : f (x, y) = c

\Rightarrow m i n o f f (x, y) a t {\begin{cases} x = \pm 1 \\ y = 0 \end{cases} w i t h m i n (f) = - 2

∄ m a x o f f (x, y)

Answered by AST last updated on 21/Jan/24

f (x, y) : {(x^{2} - 1)}^{2} + {(2 x y)}^{2} + {(y \sqrt{2})}^{2} - 2 ⩾ - 2

(E q u a l i t y : x = \underline{+} 1, y = 0)

N o m a x i m u m, f (x, y) \to \infty a s ∣ x ∣, ∣ y ∣\to \infty

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