What is the extreme f(x,y)=4-1/9 x^2-1/16 y^2 ?

The extreme f(x,y)=4-1/9 x^2-1/16 y^2 is Maximum(0,0)

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extreme f(x,y)=4-1/9 x^2-1/16 y^2

extreme points

f (x, y) = 4 - \frac{1}{9} x^{2} - \frac{1}{16} y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{1}{8}

D (x, y) = \frac{1}{36}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = ln (\frac{1}{x + y})

f (x) = x^{2} e^{x} [- 3.1]

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

f (x) = x^{3} - 5 x^{2} + 3 x + 12

f (x) = (x - 8) e^{- 9 x}

What is the extreme f(x,y)=4-1/9 x^2-1/16 y^2 ?
The extreme f(x,y)=4-1/9 x^2-1/16 y^2 is Maximum(0,0)