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

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

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

extreme points

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

Maximum (1, 1)

Solution steps

Find the critical points:

(1, 1)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(1, 1) :

Maximum

Maximum (1, 1)

P (x, y) = 200 x^{\frac{1}{2}} y^{\frac{1}{2}}

f (x) = \frac{x ^{2}}{9 - x ^{2}}

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

190 + 8 x^{3} + x^{4}

f (x) = x (25 - 37 + 2 x) (\frac{37}{2} - x)

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