What is the extreme f(x,y)=2x^2-6xy+5y^2-x+3y+2 ?

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

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extreme f(x,y)=2x^2-6xy+5y^2-x+3y+2

extreme

Minimum (- 2, - \frac{3}{2})

Solution steps

Find the critical points:

(- 2, - \frac{3}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = 10

D (x, y) = 4

Check the sign of at each critical point

Check critical point

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

Minimum

Minimum (- 2, - \frac{3}{2})

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