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

The extreme f(x,y)=(xy)/2-x^2-y^2-2x-2y+4 is Maximum(-4/3 ,-4/3)

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

extreme points

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

Maximum (- \frac{4}{3}, - \frac{4}{3})

Solution steps

Find the critical points:

(- \frac{4}{3}, - \frac{4}{3})

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

D (x, y) = \frac{15}{4}

Check the sign of at each critical point

Check critical point

(- \frac{4}{3}, - \frac{4}{3}) :

Maximum

Maximum (- \frac{4}{3}, - \frac{4}{3})

f (x, y) = 2 x^{2} + 3 y^{2} - 4 x - 3 y - 10

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

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

(x^{4} - 4 x^{3}), - 1 \leq x \leq 4

f (x) = x - 2 cos (x)

What is the extreme f(x,y)=(xy)/2-x^2-y^2-2x-2y+4 ?
The extreme f(x,y)=(xy)/2-x^2-y^2-2x-2y+4 is Maximum(-4/3 ,-4/3)