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

The extreme f(x)=x^2+y^2-5x-4y+xy is Minimum(2,1)

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

extreme points

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

Minimum (2, 1)

Solution steps

Find the critical points:

(2, 1)

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(2, 1) :

Minimum

Minimum (2, 1)

f (x) = 6 x^{2} - 8 x^{4}

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

1 + \frac{7}{x} - \frac{4}{x ^{2}}

x x^{2} + 9

f (x) = \frac{x ^{2} - 4}{x - 1} y

What is the extreme f(x)=x^2+y^2-5x-4y+xy ?
The extreme f(x)=x^2+y^2-5x-4y+xy is Minimum(2,1)