What is the extreme f(x,y)=-2x^3-2y^3+6xy+10 ?

The extreme f(x,y)=-2x^3-2y^3+6xy+10 is Saddle(0,0),Maximum(1,1)

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extreme f(x,y)=-2x^3-2y^3+6xy+10

extreme points

f (x, y) = - 2 x^{3} - 2 y^{3} + 6 x y + 10

Saddle (0, 0), Maximum (1, 1)

Solution steps

Find the critical points:

(0, 0), (1, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 12 y

D (x, y) = 144 x y - 36

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(1, 1) :

Maximum

Saddle (0, 0), Maximum (1, 1)

ln (x) + ln (2 - x)

cos^{2} (x) - 2 sin (x)

f (x) = x^{\frac{2}{3}} \cdot (6 - x)^{\frac{1}{3}}

f (x) = x^{2} + 2 y^{2} - 9 x - 12 y

f (x) = sin (\frac{x}{2})

What is the extreme f(x,y)=-2x^3-2y^3+6xy+10 ?
The extreme f(x,y)=-2x^3-2y^3+6xy+10 is Saddle(0,0),Maximum(1,1)