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

The extreme f(x,y)=3x^2-2xy+y^2-8y is Minimum(2,6)

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

extreme points

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

Minimum (2, 6)

Solution steps

Find the critical points:

(2, 6)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(2, 6) :

Minimum

Minimum (2, 6)

f (x, y) = x^{2} + y^{2} - 6 y + 9

f (x) = x^{2}, - 1 \leq x \leq 0

3 x^{2} - 5 x + 2

f (x) = x^{4} - 4 x^{3} + 16 x + 4

f (x) = x^{7} - 7 x

What is the extreme f(x,y)=3x^2-2xy+y^2-8y ?
The extreme f(x,y)=3x^2-2xy+y^2-8y is Minimum(2,6)