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

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

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

extreme points

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

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

f (x) = \frac{( 7 ln ( x ) + 4 )}{x}

f (x) = \frac{24}{x} + 4 π r^{2}

f (x) = x^{3} - 36 x^{2} + 324 x - 3280.5

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