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

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

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

extreme points

f (x) = 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, y) = 48 x - 48 y - 6 x^{2} + 6 y^{2}

\frac{x ^{2} y}{2} - x^{2} + x y - 2 x - y^{2}

y = x^{2} - 2 x

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 5, 0 \leq x \leq 3

- x^{2} + 3 x - 2, 1 \leq x \leq 3

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