What is the extreme f(x,y)=sqrt(4+2x-4y-x^2-y^2) ?

The extreme f(x,y)=sqrt(4+2x-4y-x^2-y^2) is Maximum(1,-2)

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extreme f(x,y)=sqrt(4+2x-4y-x^2-y^2)

extreme points

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

Maximum (1, - 2)

Solution steps

Find the critical points:

(1, - 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{- x ^{2} + 2 x + 8}{( - x ^{2} + 2 x + 4 - y ^{2} - 4 y ) - y ^{2} - 4 y + 2 x + 4 - x ^{2}}

D (x, y) = \frac{9}{( - x ^{2} - y ^{2} + 4 + 2 x - 4 y ) ^{2}}

Check the sign of at each critical point

Check critical point

(1, - 2) :

Maximum

Maximum (1, - 2)

What is the extreme f(x,y)=sqrt(4+2x-4y-x^2-y^2) ?
The extreme f(x,y)=sqrt(4+2x-4y-x^2-y^2) is Maximum(1,-2)