What is the extreme f(x)=x^2+xy+y^2-34y+385 ?

The extreme f(x)=x^2+xy+y^2-34y+385 is Minimum(-34/3 , 68/3)

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extreme f(x)=x^2+xy+y^2-34y+385

extreme points

f (x) = x^{2} + x y + y^{2} - 34 y + 385

Minimum (- \frac{34}{3}, \frac{68}{3})

Solution steps

Find the critical points:

(- \frac{34}{3}, \frac{68}{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{34}{3}, \frac{68}{3}) :

Minimum

Minimum (- \frac{34}{3}, \frac{68}{3})

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

f (x) = x^{3} - \frac{3}{2} x^{2}, - 5 \leq x \leq 4

f (x) = x^{3} - \frac{3}{2} x^{2}, - 5 \leq x \leq 2

f (x) = lo g_{10} (x^{2} + 1)

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

What is the extreme f(x)=x^2+xy+y^2-34y+385 ?
The extreme f(x)=x^2+xy+y^2-34y+385 is Minimum(-34/3 , 68/3)