What is the extreme f(x,y)=x^3+y^2-2xy+7x-8y+2 ?

The extreme f(x,y)=x^3+y^2-2xy+7x-8y+2 is Minimum(1,5),Saddle(-1/3 , 11/3)

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extreme f(x,y)=x^3+y^2-2xy+7x-8y+2

extreme points

f (x, y) = x^{3} + y^{2} - 2 x y + 7 x - 8 y + 2

Minimum (1, 5), Saddle (- \frac{1}{3}, \frac{11}{3})

Solution steps

Find the critical points:

(1, 5), (- \frac{1}{3}, \frac{11}{3})

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

D (x, y) = 12 x - 4

Check the sign of at each critical point

Check critical point

(1, 5) :

Minimum

Check critical point

(- \frac{1}{3}, \frac{11}{3}) :

Saddle

Minimum (1, 5), Saddle (- \frac{1}{3}, \frac{11}{3})

f (x) = (x^{2} + 11) (9 - x^{2})

4 L W

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

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

f (x) = - x^{4} + 24 x^{2}

What is the extreme f(x,y)=x^3+y^2-2xy+7x-8y+2 ?
The extreme f(x,y)=x^3+y^2-2xy+7x-8y+2 is Minimum(1,5),Saddle(-1/3 , 11/3)