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

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

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extreme f(x,y)=2x^2+y^2-2xy+5x-3y+1

extreme points

f (x, y) = 2 x^{2} + y^{2} - 2 x y + 5 x - 3 y + 1

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) = \frac{x ^{2} + 2}{x ^{2} - 4 x}

f (x) = x^{5} - 10 x^{4}

f (x) = x^{7} ln (x)

f (x) = \frac{1}{3} x^{3} + \frac{11}{2} x^{2} + 24 x + 1

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

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