What is the extreme f(x,y)=10+2x^2-2xy+y^2+5x-3y ?

The extreme f(x,y)=10+2x^2-2xy+y^2+5x-3y is Minimum(-1, 1/2)

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extreme f(x,y)=10+2x^2-2xy+y^2+5x-3y

extreme points

f (x, y) = 10 + 2 x^{2} - 2 x y + y^{2} + 5 x - 3 y

Minimum (- 1, \frac{1}{2})

Solution steps

Find the critical points:

(- 1, \frac{1}{2})

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(- 1, \frac{1}{2}) :

Minimum

Minimum (- 1, \frac{1}{2})

f (x) = x - 4 x

f (x) = x^{4} - \frac{8}{3} x^{3} + 10

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

x^{2} - y^{2}

f (x, y) = ln (x + y + 1)

What is the extreme f(x,y)=10+2x^2-2xy+y^2+5x-3y ?
The extreme f(x,y)=10+2x^2-2xy+y^2+5x-3y is Minimum(-1, 1/2)