What is the extreme f(x,y)=e^{6x^2+4y^2+10} ?

The extreme f(x,y)=e^{6x^2+4y^2+10} is Minimum(0,0)

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extreme f(x,y)=e^{6x^2+4y^2+10}

extreme points

f (x, y) = e^{6 x^{2} + 4 y^{2} + 10}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 8 (8 e^{6 x^{2} + 4 y^{2} + 10} y^{2} + e^{6 x^{2} + 4 y^{2} + 10})

D (x, y) = 96 (12 e^{6 x^{2} + 4 y^{2} + 10} x^{2} + e^{6 x^{2} + 4 y^{2} + 10}) (8 e^{6 x^{2} + 4 y^{2} + 10} y^{2} + e^{6 x^{2} + 4 y^{2} + 10}) - 9216 e^{2 (6 x^{2} + 4 y^{2} + 10)} 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 - 8 x - y^{2} + 12 x + 160

f (x, y) = x^{2} + 3 y^{2} - y^{3} - 4 x

f (x) = 0.4 x^{2} + 110.2 x, 0 \leq x

f (x) = \frac{( 3 x )}{( 9 - x ^{2} )}, - 3 \leq x \leq 2

f (x) = \frac{1}{4} x + \frac{1}{x}, x > 0

What is the extreme f(x,y)=e^{6x^2+4y^2+10} ?
The extreme f(x,y)=e^{6x^2+4y^2+10} is Minimum(0,0)