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

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

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

extreme points

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

y = \frac{1}{5} x^{5} - \frac{8}{3} x^{3} + 16 x

f (x) = \frac{( x ^{2} - 7 )}{x - 4}

\frac{x ^{2} + 1}{x ^{2} - 1}

f (x) = \frac{x ^{2} - 12}{x - 4}

(\frac{x ^{2}}{4} - \frac{x}{2} - \frac{1}{4})^{2}

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