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

The extreme f(x,y)=-1x^2-1y^2 is Maximum(0,0)

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extreme f(x,y)=-1x^2-1y^2

extreme points

f (x, y) = - 1 x^{2} - 1 y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x, y) = 3 x^{2} - x \cdot y - y - 4 \cdot x - 1

f (x) = \frac{x ^{3} - 4 x - 5}{x + 5}, - 1 \leq x \leq 5

f (x, y) = x y (30 - x - y)

y = \frac{5 x}{x ^{2} - 1}

x^{2} - 9 x + 5

What is the extreme f(x,y)=-1x^2-1y^2 ?
The extreme f(x,y)=-1x^2-1y^2 is Maximum(0,0)