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

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

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

extreme points

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

Minimum (0, 2)

Solution steps

Find the critical points:

(0, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 e^{- 4 y + x^{2} + y^{2}} (2 y^{2} - 8 y + 9)

D (x, y) = 4 e^{- 4 y + x^{2} + y^{2}} (2 e^{x^{2} + y^{2} - 4 y} x^{2} + e^{x^{2} + y^{2} - 4 y}) (2 y^{2} - 8 y + 9) - 4 e^{2 (x^{2} + y^{2} - 4 y)} x^{2} (2 y - 4)^{2}

Check the sign of at each critical point

Check critical point

(0, 2) :

Minimum

Minimum (0, 2)

f (x) = - 5 x^{4} + 15 x^{3} - 17

f (x) = x^{3} + y^{3} + 2 x^{2} + 4 y^{2} + 6

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

f (x) = 1296 x - 0.12 x^{3}

f (x) = ln (x^{2} + 3 x + 7)

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