What is the extreme f(x,y)=x^2-3xy+6y^2+5x-2y+8 ?

The extreme f(x,y)=x^2-3xy+6y^2+5x-2y+8 is Minimum(-18/5 ,-11/15)

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extreme f(x,y)=x^2-3xy+6y^2+5x-2y+8

extreme points

f (x, y) = x^{2} - 3 x y + 6 y^{2} + 5 x - 2 y + 8

Minimum (- \frac{18}{5}, - \frac{11}{15})

Solution steps

Find the critical points:

(- \frac{18}{5}, - \frac{11}{15})

\frac{\partial ^{2} f}{\partial x ^{2}} = 12

D (x, y) = 15

Check the sign of at each critical point

Check critical point

(- \frac{18}{5}, - \frac{11}{15}) :

Minimum

Minimum (- \frac{18}{5}, - \frac{11}{15})

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

f (x) = cos (x) - sin (x)

f (x, y) = x^{3} + y^{3} - 9 x y

x^{3} - 12 x + 1

f (x) = \frac{1}{x}, - 2 \leq x \leq 3

What is the extreme f(x,y)=x^2-3xy+6y^2+5x-2y+8 ?
The extreme f(x,y)=x^2-3xy+6y^2+5x-2y+8 is Minimum(-18/5 ,-11/15)