What is the extreme f(x,y)=x^2+xy+y^2-28y+261 ?

The extreme f(x,y)=x^2+xy+y^2-28y+261 is Minimum(-28/3 , 56/3)

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extreme f(x,y)=x^2+xy+y^2-28y+261

extreme points

f (x, y) = x^{2} + x y + y^{2} - 28 y + 261

Minimum (- \frac{28}{3}, \frac{56}{3})

Solution steps

Find the critical points:

(- \frac{28}{3}, \frac{56}{3})

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(- \frac{28}{3}, \frac{56}{3}) :

Minimum

Minimum (- \frac{28}{3}, \frac{56}{3})

(x - 2) (x - 3)^{2}

f (x) = 24 x - 2 x^{3}

f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} + 5 x - 2 y

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

f (x) = x^{4} + 8 x^{3} + 22 x^{2} + 24 x + 9

What is the extreme f(x,y)=x^2+xy+y^2-28y+261 ?
The extreme f(x,y)=x^2+xy+y^2-28y+261 is Minimum(-28/3 , 56/3)