What is the extreme f(x,y)=e^{y^2+x^3-6xy-6x+2y} ?

The extreme f(x,y)=e^{y^2+x^3-6xy-6x+2y} is Minimum(6,17),Saddle(0,-1)

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extreme f(x,y)=e^{y^2+x^3-6xy-6x+2y}

extreme points

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

Minimum (6, 17), Saddle (0, - 1)

Solution steps

Find the critical points:

(6, 17), (0, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 e^{2 y + x^{3} - 6 x y + y^{2} - 6 x} (18 x^{2} - 12 x y - 12 x + 2 y^{2} + 4 y + 3)

D (x, y) = 6 e^{2 x^{3} - 12 x y - 12 x + 2 y^{2} + 4 y} (3 (x^{2} - 2 y - 2)^{2} + 2 x) (18 x^{2} - 12 x y - 12 x + 2 y^{2} + 4 y + 3) - 36 e^{2 (2 y + x^{3} - 6 x y + y^{2} - 6 x)} (- 3 x^{3} + x^{2} y + x^{2} + 6 x y + 6 x - 2 y^{2} - 4 y - 3)^{2}

Check the sign of at each critical point

Check critical point

(6, 17) :

Minimum

Check critical point

(0, - 1) :

Saddle

Minimum (6, 17), Saddle (0, - 1)

\frac{2 x + 4}{x ^{2} - 16}

f (x) = x^{2} \frac{1}{x - 1}

f (x) = x [18 - 2 (x)]^{2}

f (x, y) = x^{2} - 3 y^{2} - 8 x + 9 y + 3 x y

f (d, y) = \frac{d}{d ^{2}} (y)

What is the extreme f(x,y)=e^{y^2+x^3-6xy-6x+2y} ?
The extreme f(x,y)=e^{y^2+x^3-6xy-6x+2y} is Minimum(6,17),Saddle(0,-1)