What is the extreme f(x,y)=-x^3-y^2+6xy+2 ?

The extreme f(x,y)=-x^3-y^2+6xy+2 is Saddle(0,0),Maximum(6,18)

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extreme f(x,y)=-x^3-y^2+6xy+2

extreme points

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

Saddle (0, 0), Maximum (6, 18)

Solution steps

Find the critical points:

(0, 0), (6, 18)

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

D (x, y) = 12 x - 36

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(6, 18) :

Maximum

Saddle (0, 0), Maximum (6, 18)

f (x) = (x - 1)^{3} (x - 2)^{2}

f (x) = \frac{x}{ln ( x )}, 2 \leq x \leq 5

f (x) = x^{\frac{4}{3}} - 2 x^{\frac{1}{3}}

\frac{x}{x ^{2} + 12 x + 35}

f (x, y) = 2 x y - x^{2} - 2 y^{2} + 3 x + 4

What is the extreme f(x,y)=-x^3-y^2+6xy+2 ?
The extreme f(x,y)=-x^3-y^2+6xy+2 is Saddle(0,0),Maximum(6,18)