What is the extreme f(x,y)=x^2-4xy+5y^2+2x+4y-4 ?

The extreme f(x,y)=x^2-4xy+5y^2+2x+4y-4 is Minimum(-9,-4)

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extreme f(x,y)=x^2-4xy+5y^2+2x+4y-4

extreme points

f (x, y) = x^{2} - 4 x y + 5 y^{2} + 2 x + 4 y - 4

Minimum (- 9, - 4)

Solution steps

Find the critical points:

(- 9, - 4)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(- 9, - 4) :

Minimum

Minimum (- 9, - 4)

f (x) = 1 + \frac{1}{x} + \frac{9}{x ^{2}} + \frac{1}{x ^{3}}

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

0.0025 x^{4} + 0.5 x^{3} - 8 x^{2} + 1000

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

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

What is the extreme f(x,y)=x^2-4xy+5y^2+2x+4y-4 ?
The extreme f(x,y)=x^2-4xy+5y^2+2x+4y-4 is Minimum(-9,-4)