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

The extreme f(xy)=4-x^2-2y^2 is Maximum(0,0)

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

extreme points

f (x y) = 4 - x^{2} - 2 y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x, y) = x^{3} - 12 x + y^{2} + 6 y + 14

f (x) = \frac{1}{5} x^{5} - \frac{2}{3} x^{3} + x

f (x) = \frac{1}{16} x^{3} - 4

\frac{x - 5}{4}

f (x, y) = x^{2} + y^{2} - 18 x + 16 y - 7

What is the extreme f(xy)=4-x^2-2y^2 ?
The extreme f(xy)=4-x^2-2y^2 is Maximum(0,0)