What is the extreme f(x,y)=e^{9x^2+6y^2+1} ?

The extreme f(x,y)=e^{9x^2+6y^2+1} is Minimum(0,0)

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extreme f(x,y)=e^{9x^2+6y^2+1}

extreme points

f (x, y) = e^{9 x^{2} + 6 y^{2} + 1}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 12 (12 e^{9 x^{2} + 6 y^{2} + 1} y^{2} + e^{9 x^{2} + 6 y^{2} + 1})

D (x, y) = 216 (18 e^{9 x^{2} + 6 y^{2} + 1} x^{2} + e^{9 x^{2} + 6 y^{2} + 1}) (12 e^{9 x^{2} + 6 y^{2} + 1} y^{2} + e^{9 x^{2} + 6 y^{2} + 1}) - 46656 e^{2 (9 x^{2} + 6 y^{2} + 1)} x^{2} y^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = (\frac{3 - cos ( θ )}{sin ( θ )})

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

\frac{9}{8} y \frac{3}{4}

f (x) = - 2 x^{3} + 27 x^{2} - 108 x, 2 \leq x \leq 7

f (x, y) = 64 - 4 x^{2} - y^{2}

What is the extreme f(x,y)=e^{9x^2+6y^2+1} ?
The extreme f(x,y)=e^{9x^2+6y^2+1} is Minimum(0,0)