What is the extreme f(x)=e^{3x^2+8y^2+13} ?

The extreme f(x)=e^{3x^2+8y^2+13} is Minimum(0,0)

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extreme f(x)=e^{3x^2+8y^2+13}

extreme

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 16 (16 e^{3 x^{2} + 8 y^{2} + 13} y^{2} + e^{3 x^{2} + 8 y^{2} + 13})

D (x, y) = 96 (6 e^{3 x^{2} + 8 y^{2} + 13} x^{2} + e^{3 x^{2} + 8 y^{2} + 13}) (16 e^{3 x^{2} + 8 y^{2} + 13} y^{2} + e^{3 x^{2} + 8 y^{2} + 13}) - 9216 e^{2 (3 x^{2} + 8 y^{2} + 13)} x^{2} y^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

What is the extreme f(x)=e^{3x^2+8y^2+13} ?
The extreme f(x)=e^{3x^2+8y^2+13} is Minimum(0,0)