What is the extreme f(x,y)=e^{x^2+3y^2+15} ?

The extreme f(x,y)=e^{x^2+3y^2+15} is Minimum(0,0)

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extreme f(x,y)=e^{x^2+3y^2+15}

extreme points

f (x, y) = e^{x^{2} + 3 y^{2} + 15}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 (6 e^{x^{2} + 3 y^{2} + 15} y^{2} + e^{x^{2} + 3 y^{2} + 15})

D (x, y) = 12 (2 e^{x^{2} + 3 y^{2} + 15} x^{2} + e^{x^{2} + 3 y^{2} + 15}) (6 e^{x^{2} + 3 y^{2} + 15} y^{2} + e^{x^{2} + 3 y^{2} + 15}) - 144 e^{2 (x^{2} + 3 y^{2} + 15)} x^{2} y^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

g (x) = x^{3} - 2 x^{2} - 4 x + 8

f (x) = x^{3} + 6 x^{2} - 63 x + 5, - 8 \leq x \leq 0

f (x) = 3 x^{2} - 6 x + 24

f (x) = (9 x^{2} + 4 x + 5) (9 y^{2} + 6 y + 6)

f (x) = \frac{4860}{x} + 12 x + 709407

What is the extreme f(x,y)=e^{x^2+3y^2+15} ?
The extreme f(x,y)=e^{x^2+3y^2+15} is Minimum(0,0)