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

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

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 24 (4 e^{2 x^{2} + 3 y^{2}} x^{2} + e^{2 x^{2} + 3 y^{2}}) (6 e^{2 x^{2} + 3 y^{2}} y^{2} + e^{2 x^{2} + 3 y^{2}}) - 576 e^{2 (2 x^{2} + 3 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) = 2 (3 - x) [- 1.2]

y = 6 x - x^{2}

f (x) = \frac{- 4 x + 92}{( x - 119 ) ^{7}}

f (x) = x^{\frac{1}{x}}

f (x) = 2 x + 7 x^{\frac{2}{7}}

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