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

The extreme f(x,y)=e^{-(x^2+y^2-6y)} is Maximum(0,3)

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

extreme points

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

Maximum (0, 3)

Solution steps

Find the critical points:

(0, 3)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 e^{6 y - x^{2} - y^{2}} (2 y^{2} - 12 y + 17)

D (x, y) = - 4 e^{6 y - x^{2} - y^{2}} (- 2 e^{- x^{2} - y^{2} + 6 y} x^{2} + e^{- x^{2} - y^{2} + 6 y}) (2 y^{2} - 12 y + 17) - 4 e^{2 (- x^{2} - y^{2} + 6 y)} x^{2} (- 2 y + 6)^{2}

Check the sign of at each critical point

Check critical point

(0, 3) :

Maximum

Maximum (0, 3)

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

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

E (a, b) = \frac{a ( a + b ) ^{2} ( a - b )}{a ^{2} - b ^{2}}

f (x) = (x^{2} - \frac{7}{3} x + \frac{13}{9}) e^{3 x}

f (x) = 2 x^{5} + 2 x^{4} - x^{3} + 6 x^{2} + 8 x + 4

What is the extreme f(x,y)=e^{-(x^2+y^2-6y)} ?
The extreme f(x,y)=e^{-(x^2+y^2-6y)} is Maximum(0,3)