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

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

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extreme f(x,y)=x^2+xy+y^2+3/x+3/y+2

extreme points

f (x, y) = x^{2} + x y + y^{2} + \frac{3}{x} + \frac{3}{y} + 2

Minimum (1, 1)

Solution steps

Find the critical points:

(1, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{6}{y ^{3}} + 2

D (x, y) = \frac{36}{x ^{3} y ^{3}} + \frac{12}{x ^{3}} + \frac{12}{y ^{3}} + 3

Check the sign of at each critical point

Check critical point

(1, 1) :

Minimum

Minimum (1, 1)

y = 2 + s + 2 (- 3 - 2 s + x)

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

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

f (x) = x^{4} (1 - x)^{8}

f (x) = - x e^{- \frac{x}{4}}

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