Determine the maximum and minimum of the function: f(x,y)=x^4 +4x^2 y^2 −2x^2 +2y^2 −1 Thank you


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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
	$let us see for which c∈R, ∃(x,y):f(x,y)=c x^2 =a, y^2 =b ⇒f=a^2 +4ab−2a+2b−1=c (a^2 −2a−1)+2b(2a+1)=c ⇒b=((−a^2 +2a+c+1)/(2a+1))=y^2 ≥0 also a=x^2 ≥0 −a^2 +2a+c+1=0⇒a_(1,2) =((−2±(√(8+4c)))/(−2))=1±(√(2+c)) 2a+1=0 ⇒ a_3 =−1/2 ⇒a≥0, 2a+1>0 1st case: c<−2 ⇒ −a^2 +2a+1+c<0 ⇒b<0→contradiction 2nd case: c=−2 ⇒ −a^2 +2a+1+c≤0 where for a=1 and b=0 and for a>0, b<0 → contradict 3rd case: −1≥c>−2 ⇒b≥0 for 1−(√(2+c))≤a≤1+(√(2+c)) 4th case: c>−1 ⇒b≥0 for 0≤a≤1+(√(2+c)) ⇒ for all c≥−2, ∃(x,y):f(x,y)=c ⇒min of f(x,y) at { ((x=±1)),((y=0)) :} with min(f)=−2 ∄max of f(x,y)$
	$l e t u s s e e f o r w h i c h c \in R, \exists (x, y) : f (x, y) = c$ $x^{2} = a, y^{2} = b$ $\Rightarrow f = a^{2} + 4 a b - 2 a + 2 b - 1 = c$ $(a^{2} - 2 a - 1) + 2 b (2 a + 1) = c$ $\Rightarrow b = \frac{- a^{2} + 2 a + c + 1}{2 a + 1} = y^{2} ⩾ 0$ $a l s o a = x^{2} ⩾ 0$ $- a^{2} + 2 a + c + 1 = 0 \Rightarrow a_{1, 2} = \frac{- 2 \pm \sqrt{8 + 4 c}}{- 2} = 1 \pm \sqrt{2 + c}$ $2 a + 1 = 0 \Rightarrow a_{3} = - 1 / 2$ $\Rightarrow a ⩾ 0, 2 a + 1 > 0$ $1 s t c a s e : c < - 2 \Rightarrow - a^{2} + 2 a + 1 + c < 0$ $\Rightarrow b < 0 \to c o n t r a d i c t i o n$ $2 n d c a s e : c = - 2 \Rightarrow - a^{2} + 2 a + 1 + c ⩽ 0$ $w h e r e f o r a = 1 a n d b = 0$ $a n d f o r a > 0, b < 0 \to c o n t r a d i c t$ $3 r d c a s e : - 1 ⩾ c > - 2$ $\Rightarrow b ⩾ 0 f o r 1 - \sqrt{2 + c} ⩽ a ⩽ 1 + \sqrt{2 + c}$ $4 t h c a s e : c > - 1$ $\Rightarrow b ⩾ 0 f o r 0 ⩽ a ⩽ 1 + \sqrt{2 + c}$ $\Rightarrow f o r a l l c ⩾ - 2, \exists (x, y) : f (x, y) = c$ $\Rightarrow m i n o f f (x, y) a t {\begin{cases} x = \pm 1 \\ y = 0 \end{cases} w i t h m i n (f) = - 2$ $∄ m a x o f f (x, y)$
	Answered by deleteduser1 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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