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

The extreme P(x,y)=2x+2y-x^2+y^2 is Saddle(1,-1)

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

extreme points

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

Saddle (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) :

Saddle

Saddle (1, - 1)

f (x) = x^{3} - \frac{3}{2} x^{2}, - 1 \leq x \leq 4

f (x) = 2 x^{3} + 3 x^{2} + 4, - 2 \leq x \leq 1

f (x) = x^{3} + 3 x^{2} - 24 x [0.3]

f (x) = - \frac{2}{x - 7}

f (x) = x^{2} + x + 4

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