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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, 3) :

Maximum

Maximum (0, 3)

2 x^{3} + 6 x^{2} - 48 x

f (x) = - 2 x^{2} - 20 x + 8

f (x) = x^{5} - x^{3} + 1

y = \frac{x}{x ^{2} + 16}

- x^{3} (x - 3)

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)