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

The extreme f(x,y)=e^{2y-x^2-y^2} is Maximum(0,1)

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

extreme points

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

Maximum (0, 1)

Solution steps

Find the critical points:

(0, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 e^{2 y - x^{2} - y^{2}} (2 y^{2} - 4 y + 1)

D (x, y) = - 4 e^{2 y - x^{2} - y^{2}} (- 2 e^{2 y - x^{2} - y^{2}} x^{2} + e^{2 y - x^{2} - y^{2}}) (2 y^{2} - 4 y + 1) - 4 e^{2 (2 y - x^{2} - y^{2})} x^{2} (- 2 y + 2)^{2}

Check the sign of at each critical point

Check critical point

(0, 1) :

Maximum

Maximum (0, 1)

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

f (x) = - 4 x^{3} + 12 x^{2} - 10

f (x) = 2 x + 7 ln (x)

68 - 11

f (x) = x^{3} - 75 x + 3

What is the extreme f(x,y)=e^{2y-x^2-y^2} ?
The extreme f(x,y)=e^{2y-x^2-y^2} is Maximum(0,1)