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

The extreme f(x,y)=4xy+y^2 is Saddle(0,0)

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extreme f(x,y)=4xy+y^2

extreme points

f (x, y) = 4 x y + y^{2}

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

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

f (x) = \frac{( 1 )}{( 3 ) x ^{(3)} - 9}

f (x) = \frac{x ^{3}}{3} - 65 x^{2} + 4000 x - 100

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

f (x) = - 8 + 4 x^{2}