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

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

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

extreme points

f (x, y) = x^{2} - 3 x y + 5 x - 2 y + 6 y^{2} + 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) = \frac{3}{4} x^{4} - 3 x^{3} - 81 x^{2} - 16

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

f (x) = x^{2} - 5

f (x, y) = 5 e^{7 x y}

f (x) = 8 ln (x) - x^{2}

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