What is the extreme f(x)=2x^2+3xy+4y^2-2x+10y ?

The extreme f(x)=2x^2+3xy+4y^2-2x+10y is Minimum(2,-2)

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extreme f(x)=2x^2+3xy+4y^2-2x+10y

extreme points

f (x) = 2 x^{2} + 3 x y + 4 y^{2} - 2 x + 10 y

Minimum (2, - 2)

Solution steps

Find the critical points:

(2, - 2)

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

D (x, y) = 23

Check the sign of at each critical point

Check critical point

(2, - 2) :

Minimum

Minimum (2, - 2)

What is the extreme f(x)=2x^2+3xy+4y^2-2x+10y ?
The extreme f(x)=2x^2+3xy+4y^2-2x+10y is Minimum(2,-2)