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

Commented bybemath last updated on 22/Sep/20

santuyy...sir. gave kudos