What is the extreme f(x,y)=x^3+y^3+7xy ?

The extreme f(x,y)=x^3+y^3+7xy is Saddle(0,0),Maximum(-7/3 ,-7/3)

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extreme f(x,y)=x^3+y^3+7xy

extreme points

f (x, y) = x^{3} + y^{3} + 7 x y

Saddle (0, 0), Maximum (- \frac{7}{3}, - \frac{7}{3})

Solution steps

Find the critical points:

(0, 0), (- \frac{7}{3}, - \frac{7}{3})

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y

D (x, y) = 36 x y - 49

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- \frac{7}{3}, - \frac{7}{3}) :

Maximum

Saddle (0, 0), Maximum (- \frac{7}{3}, - \frac{7}{3})

f (x) = 59.80 8^{2} - 11.148 x

f (x) = \frac{x ^{2} + 2}{x ^{2} - 16}

f (x, y) = x^{3} + 12 x y + y^{4}

f (x) = x^{3} e^{2 x} + 1

What is the extreme f(x,y)=x^3+y^3+7xy ?
The extreme f(x,y)=x^3+y^3+7xy is Saddle(0,0),Maximum(-7/3 ,-7/3)