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

The extreme f(x)=x^2+y^2+x+y+xy is Minimum(-1/3 ,-1/3)

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extreme f(x)=x^2+y^2+x+y+xy

extreme points

f (x) = x^{2} + y^{2} + x + y + x y

Minimum (- \frac{1}{3}, - \frac{1}{3})

Solution steps

Find the critical points:

(- \frac{1}{3}, - \frac{1}{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{1}{3}, - \frac{1}{3}) :

Minimum

Minimum (- \frac{1}{3}, - \frac{1}{3})

f (x) = \frac{ln ( x )}{8 x}, 1 \leq x \leq 4

f (x) = \frac{1 + 3 x}{4 + 5 x ^{2}}

f (x) = x^{2} + y^{2} - 6 x + 3 y

f (x) = (2 x^{2} + 1)^{2}

f (x, y) = \frac{100}{1 + x ^{2} + y ^{2}}

What is the extreme f(x)=x^2+y^2+x+y+xy ?
The extreme f(x)=x^2+y^2+x+y+xy is Minimum(-1/3 ,-1/3)