What is the extreme f(x,y)=3e^{x^2+y^2} ?

The extreme f(x,y)=3e^{x^2+y^2} is Minimum(0,0)

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 (2 e^{x^{2} + y^{2}} y^{2} + e^{x^{2} + y^{2}})

D (x, y) = 36 (2 e^{x^{2} + y^{2}} x^{2} + e^{x^{2} + y^{2}}) (2 e^{x^{2} + y^{2}} y^{2} + e^{x^{2} + y^{2}}) - 144 e^{2 (x^{2} + y^{2})} x^{2} y^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x, y) = (x - y) (2 x + y) (x + 1)

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

2 x^{2} + 9

y = 2 x^{4} - 12 x^{2}

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

What is the extreme f(x,y)=3e^{x^2+y^2} ?
The extreme f(x,y)=3e^{x^2+y^2} is Minimum(0,0)