What is the extreme f(x)=x^2+xy+12y^2-4x+y ?

The extreme f(x)=x^2+xy+12y^2-4x+y is Minimum(97/47 ,-6/47)

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extreme f(x)=x^2+xy+12y^2-4x+y

extreme points

f (x) = x^{2} + x y + 12 y^{2} - 4 x + y

Minimum (\frac{97}{47}, - \frac{6}{47})

Solution steps

Find the critical points:

(\frac{97}{47}, - \frac{6}{47})

\frac{\partial ^{2} f}{\partial x ^{2}} = 24

D (x, y) = 47

Check the sign of at each critical point

Check critical point

(\frac{97}{47}, - \frac{6}{47}) :

Minimum

Minimum (\frac{97}{47}, - \frac{6}{47})

f (x, y) = 6 x + 6 x y + y

f (x) = (x t)

- (y - 3) (y + 5)

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

f (x, y) = x y - 7 y - 49 x + 343

What is the extreme f(x)=x^2+xy+12y^2-4x+y ?
The extreme f(x)=x^2+xy+12y^2-4x+y is Minimum(97/47 ,-6/47)