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

The extreme f(x,y)=-x^2+x-y^2-2y is Maximum(1/2 ,-1)

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extreme f(x,y)=-x^2+x-y^2-2y

extreme points

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

Maximum (\frac{1}{2}, - 1)

Solution steps

Find the critical points:

(\frac{1}{2}, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 2

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(\frac{1}{2}, - 1) :

Maximum

Maximum (\frac{1}{2}, - 1)

f (x) = 8 x ln (x)

f (x) = x^{3} - 3 x^{2} - 9 x + 11

f (x) = x^{3} - 3 x^{2} - 9 x + 20

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

f (x, y) = x^{3} + y^{3} - 3 x y + 4

What is the extreme f(x,y)=-x^2+x-y^2-2y ?
The extreme f(x,y)=-x^2+x-y^2-2y is Maximum(1/2 ,-1)