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

The extreme f(x,y)=x^2+xy+1/2 y^2-5x+y is Minimum(6,-7)

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

extreme points

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

Minimum (6, - 7)

Solution steps

Find the critical points:

(6, - 7)

\frac{\partial ^{2} f}{\partial x ^{2}} = 1

D (x, y) = 1

Check the sign of at each critical point

Check critical point

(6, - 7) :

Minimum

Minimum (6, - 7)

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

(x^{2} + 3 x - 1)^{\frac{1}{3}}

f (x, y) = 5 x^{2} + 5 y^{2} + 20 x - 10 y + 40

f (x) = x (27 - 41 + 2 x) (\frac{41}{2} - x)

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

What is the extreme f(x,y)=x^2+xy+1/2 y^2-5x+y ?
The extreme f(x,y)=x^2+xy+1/2 y^2-5x+y is Minimum(6,-7)