What is the extreme f(x,y)=10-(x^2)/2-(y^2)/2 ?

The extreme f(x,y)=10-(x^2)/2-(y^2)/2 is Maximum(0,0)

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extreme f(x,y)=10-(x^2)/2-(y^2)/2

extreme points

f (x, y) = 10 - \frac{x ^{2}}{2} - \frac{y ^{2}}{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 1

D (x, y) = 1

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

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

f (x, y) = 6 y^{3} - 5 x^{3} y + 3 x^{2}

f (x) = - \frac{x ^{2}}{2} + \frac{x}{3}

f (x) = (x - 1)^{2} \cdot (x - 3) + 2

f (x) = 14 x + 19 x^{\frac{14}{19}}

What is the extreme f(x,y)=10-(x^2)/2-(y^2)/2 ?
The extreme f(x,y)=10-(x^2)/2-(y^2)/2 is Maximum(0,0)