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

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

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

extreme points

f (x) = 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) = x^{3} + 9 x^{2} + y^{2} + 24 x + 2 y + 17

f (x) = \frac{2 x - 5}{x + 6}

2^{x^{2} - 5}

f (x) = x^{\frac{3}{4}} - 4 x^{\frac{1}{4}}

x = \frac{2}{x}

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