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

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

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

extreme points

f (x, y) = 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})

(x - 1)^{3} + 2

f (x) = x^{5} \cdot ln (x)

f (x) = ((42 - x) x - (2 x + 80))

f (x, y) = - 4 x + y - \frac{1}{x} + \frac{1}{y}

f (x) = \frac{x ^{2} + 4}{x ^{2} - 25}

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