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Question Number 114945 by bemath last updated on 22/Sep/20

	Answered by bobhans last updated on 22/Sep/20
	$f(x,y) = x^3 +y^3 −3x−12y+20 f_x =3x^2 −3=0 → { ((x=1)),((x=−1)) :} f_y =3y^2 −12=0→ { ((y=2)),((y=−2)) :} f_(xx) = 6x ; f_(yy) = 6y ; f_(xy) =f_(yx) =0 the function possible extremum at (1,2); (1,−2);(−1,2) and (−1,−2) next D is evaluated at this four points D=f_(xx) .f_(yy) −(f_(xy) )^2 =36xy at { (((1,2) →D>0 ∧f_(xx) >0 , local min at(1,2))),(((−1,2)→D<0 has no extremum at(−1,2))) :} at { (((−1,−2)⇒D>0 ∧f_(xx) <0 ,local max at (−1,−2))),(((1,−2)→D<0 has no extremum at (1,−2))) :} min value f(1,2)=1^3 +2^3 −3.1−12.2+20 = 29−27=2 max value f(−1,−2)=−1−8+3+24+20 = −9+47=38$
	$f (x, y) = x^{3} + y^{3} - 3 x - 12 y + 20$ $f_{x} = 3 x^{2} - 3 = 0 \to {\begin{cases} x = 1 \\ x = - 1 \end{cases}$ $f_{y} = 3 y^{2} - 12 = 0 \to {\begin{cases} y = 2 \\ y = - 2 \end{cases}$ $f_{x x} = 6 x; f_{y y} = 6 y; f_{x y} = f_{y x} = 0$ $t h e f u n c t i o n p o s s i b l e e x t r e m u m a t$ $(1, 2); (1, - 2); (- 1, 2) a n d (- 1, - 2)$ $n e x t D i s e v a l u a t e d a t t h i s f o u r p o i n t s$ $D = f_{x x} . f_{y y} - {(f_{x y})}^{2} = 36 x y$ $a t {\begin{cases} (1, 2) \to D > 0 \land f_{x x} > 0, l o c a l m i n a t (1, 2) \\ (- 1, 2) \to D < 0 h a s n o e x t r e m u m a t (- 1, 2) \end{cases}$ $a t {\begin{cases} (- 1, - 2) \Rightarrow D > 0 \land f_{x x} < 0, l o c a l m a x a t (- 1, - 2) \\ (1, - 2) \to D < 0 h a s n o e x t r e m u m a t (1, - 2) \end{cases}$ $m i n v a l u e f (1, 2) = 1^{3} + 2^{3} - 3.1 - 12.2 + 20$ $= 29 - 27 = 2$ $m a x v a l u e f (- 1, - 2) = - 1 - 8 + 3 + 24 + 20$ $= - 9 + 47 = 38$
	Commented by bemath last updated on 22/Sep/20

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