What is the extreme f(x,y)=y^3+x^2-6xy+3x+6y-7 ?

The extreme f(x,y)=y^3+x^2-6xy+3x+6y-7 is Minimum(27/2 ,5),Saddle(3/2 ,1)

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extreme f(x,y)=y^3+x^2-6xy+3x+6y-7

extreme points

f (x, y) = y^{3} + x^{2} - 6 x y + 3 x + 6 y - 7

Minimum (\frac{27}{2}, 5), Saddle (\frac{3}{2}, 1)

Solution steps

Find the critical points:

(\frac{27}{2}, 5), (\frac{3}{2}, 1)

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

D (x, y) = 12 y - 36

Check the sign of at each critical point

Check critical point

(\frac{27}{2}, 5) :

Minimum

Check critical point

(\frac{3}{2}, 1) :

Saddle

Minimum (\frac{27}{2}, 5), Saddle (\frac{3}{2}, 1)

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

x^{2} - 4

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

K (a, b) = 4 a + 9 ab

e^{- 2.5 x^{2}}

What is the extreme f(x,y)=y^3+x^2-6xy+3x+6y-7 ?
The extreme f(x,y)=y^3+x^2-6xy+3x+6y-7 is Minimum(27/2 ,5),Saddle(3/2 ,1)