What is the extreme f(-7/3 , 14/3)=x^2+xy+y^2-7y+16 ?

The extreme f(-7/3 , 14/3)=x^2+xy+y^2-7y+16 is Minimum(-7/3 , 14/3)

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extreme f(-7/3 , 14/3)=x^2+xy+y^2-7y+16

extreme points

f (- \frac{7}{3}, \frac{14}{3}) = x^{2} + x y + y^{2} - 7 y + 16

Minimum (- \frac{7}{3}, \frac{14}{3})

Solution steps

Find the critical points:

(- \frac{7}{3}, \frac{14}{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{7}{3}, \frac{14}{3}) :

Minimum

Minimum (- \frac{7}{3}, \frac{14}{3})

f (x, y) = 3 x^{2} - 9 x + 9 x y^{2}

2 θ - 4 sin (θ)

f (1.1, 1.1) = 2 x + 3 y - x^{3} - 2 y^{2}

f (x) = 9 - x

f (x) = x^{9} e^{2 x}

What is the extreme f(-7/3 , 14/3)=x^2+xy+y^2-7y+16 ?
The extreme f(-7/3 , 14/3)=x^2+xy+y^2-7y+16 is Minimum(-7/3 , 14/3)