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

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

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

extreme points

f (x) = 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})

\frac{- x + 3}{x ^{2} + 14 x + 49}

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

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

f (x) = - 2 cos (2 x)

f (x) = 3 x, - 8 \leq x \leq 8

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