What is the extreme f(x,y)=9e^{4y-x^2-y^2} ?

The extreme f(x,y)=9e^{4y-x^2-y^2} is Maximum(0,2)

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extreme f(x,y)=9e^{4y-x^2-y^2}

extreme points

f (x, y) = 9 e^{4 y - x^{2} - y^{2}}

Maximum (0, 2)

Solution steps

Find the critical points:

(0, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 9 (4 e^{4 y - x^{2} - y^{2}} y^{2} - 16 e^{4 y - x^{2} - y^{2}} y + 14 e^{4 y - x^{2} - y^{2}})

D (x, y) = - 162 (- 2 e^{4 y - x^{2} - y^{2}} x^{2} + e^{4 y - x^{2} - y^{2}}) (4 e^{4 y - x^{2} - y^{2}} y^{2} - 16 e^{4 y - x^{2} - y^{2}} y + 14 e^{4 y - x^{2} - y^{2}}) - 324 e^{2 (4 y - x^{2} - y^{2})} x^{2} (- 2 y + 4)^{2}

Check the sign of at each critical point

Check critical point

(0, 2) :

Maximum

Maximum (0, 2)

f (x) = x - r x (1 - x)

f (x, y) = x^{3} + y^{3} + 3 x^{2} - 6 y^{2} - 5

f (x) = 0.01 x^{3} - 0.45 x^{2} + 2.43 x + 300

f (x) = (x - 2)^{3} + x^{2} - 1

f (x) = \frac{1}{2} x^{4} - 4 x^{3} + 8 x^{2} + 46

What is the extreme f(x,y)=9e^{4y-x^2-y^2} ?
The extreme f(x,y)=9e^{4y-x^2-y^2} is Maximum(0,2)