What is the extreme f(x,y)=x^2+3xy+4y^2-6x+2y ?

The extreme f(x,y)=x^2+3xy+4y^2-6x+2y is Minimum(54/7 ,-22/7)

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extreme f(x,y)=x^2+3xy+4y^2-6x+2y

extreme points

f (x, y) = x^{2} + 3 x y + 4 y^{2} - 6 x + 2 y

Minimum (\frac{54}{7}, - \frac{22}{7})

Solution steps

Find the critical points:

(\frac{54}{7}, - \frac{22}{7})

\frac{\partial ^{2} f}{\partial x ^{2}} = 8

D (x, y) = 7

Check the sign of at each critical point

Check critical point

(\frac{54}{7}, - \frac{22}{7}) :

Minimum

Minimum (\frac{54}{7}, - \frac{22}{7})

What is the extreme f(x,y)=x^2+3xy+4y^2-6x+2y ?
The extreme f(x,y)=x^2+3xy+4y^2-6x+2y is Minimum(54/7 ,-22/7)