What is the extreme f(x,y)=x^3+y^3-63(x+y)+12xy ?

The extreme f(x,y)=x^3+y^3-63(x+y)+12xy is Saddle(-1,5),Minimum(3,3),Saddle(5,-1),Maximum(-7,-7)

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extreme f(x,y)=x^3+y^3-63(x+y)+12xy

extreme points

f (x, y) = x^{3} + y^{3} - 63 (x + y) + 12 x y

Saddle (- 1, 5), Minimum (3, 3), Saddle (5, - 1), Maximum (- 7, - 7)

Solution steps

Find the critical points:

(- 1, 5), (3, 3), (5, - 1), (- 7, - 7)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y

D (x, y) = 36 x y - 144

Check the sign of at each critical point

Check critical point

(- 1, 5) :

Saddle

Check critical point

(3, 3) :

Minimum

Check critical point

(5, - 1) :

Saddle

Check critical point

(- 7, - 7) :

Maximum

Saddle (- 1, 5), Minimum (3, 3), Saddle (5, - 1), Maximum (- 7, - 7)

f (x, y) = x^{2} + 3 y^{2} - 4 x + 6 y

g (x) = \frac{1}{9} x^{9} - x

f (x, y) = x^{2} + y^{2} - 18 x + 12 y - 7

f (x) = x^{2} - 6 x + 6

f (x) = 2 x^{2} - 8 x + 11, 0 \leq x \leq 6

What is the extreme f(x,y)=x^3+y^3-63(x+y)+12xy ?
The extreme f(x,y)=x^3+y^3-63(x+y)+12xy is Saddle(-1,5),Minimum(3,3),Saddle(5,-1),Maximum(-7,-7)