What is the extreme f(x)=x^2+xy+y^2-10y+33 ?

The extreme f(x)=x^2+xy+y^2-10y+33 is Minimum(-10/3 , 20/3)

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extreme f(x)=x^2+xy+y^2-10y+33

extreme points

f (x) = x^{2} + x y + y^{2} - 10 y + 33

Minimum (- \frac{10}{3}, \frac{20}{3})

Solution steps

Find the critical points:

(- \frac{10}{3}, \frac{20}{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{10}{3}, \frac{20}{3}) :

Minimum

Minimum (- \frac{10}{3}, \frac{20}{3})

f (x) = x^{3} - 6 x^{2} + 12 x + 1

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

f (x) = e^{5 x} + e^{- x}

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

x^{5} - 15 x^{3}

What is the extreme f(x)=x^2+xy+y^2-10y+33 ?
The extreme f(x)=x^2+xy+y^2-10y+33 is Minimum(-10/3 , 20/3)