What is the extreme f(x,y)=x^2+xy+y^2-25y+208 ?

The extreme f(x,y)=x^2+xy+y^2-25y+208 is Minimum(-25/3 , 50/3)

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extreme f(x,y)=x^2+xy+y^2-25y+208

extreme points

f (x, y) = x^{2} + x y + y^{2} - 25 y + 208

Minimum (- \frac{25}{3}, \frac{50}{3})

Solution steps

Find the critical points:

(- \frac{25}{3}, \frac{50}{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{25}{3}, \frac{50}{3}) :

Minimum

Minimum (- \frac{25}{3}, \frac{50}{3})

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

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

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

f (t) = u (t)

f (x) = 3 x^{4} + 4 x^{3} - 36 x^{2} - 78

What is the extreme f(x,y)=x^2+xy+y^2-25y+208 ?
The extreme f(x,y)=x^2+xy+y^2-25y+208 is Minimum(-25/3 , 50/3)