What is the extreme f(x,y)=18xy+2500 ?

The extreme f(x,y)=18xy+2500 is Saddle(0,0)

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extreme f(x,y)=18xy+2500

extreme points

f (x, y) = 18 x y + 2500

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, y) = - 324

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = - (x + 2)^{2} + 8

f (x) = I x + 1 I - I x + 1 I

y = x^{3} - 6 x^{2} + 9 x + 5

e^{8 x^{2} + 5 x^{2} + 8}

f (x) = \frac{1}{x ^{2} + 2 x + 9}, - 2 \leq x \leq 1

What is the extreme f(x,y)=18xy+2500 ?
The extreme f(x,y)=18xy+2500 is Saddle(0,0)