What is the extreme f(x,y)=2x^3+6xy^2-3y^3-150x ?

The extreme f(x,y)=2x^3+6xy^2-3y^3-150x is Saddle(3,4),Saddle(-3,-4)

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extreme f(x,y)=2x^3+6xy^2-3y^3-150x

extreme points

f (x, y) = 2 x^{3} + 6 x y^{2} - 3 y^{3} - 150 x

Saddle (3, 4), Saddle (- 3, - 4)

Solution steps

Find the critical points:

(3, 4), (- 3, - 4)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 18 y + 12 x

D (x, y) = 12 x (- 18 y + 12 x) - 144 y^{2}

Check the sign of at each critical point

Check critical point

(3, 4) :

Saddle

Check critical point

(- 3, - 4) :

Saddle

Saddle (3, 4), Saddle (- 3, - 4)

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

f (x) = 12 x + ln (x)

ln (27 + 8 x^{3})

x^{3} - 6 x^{2} - 15 x + 7

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

What is the extreme f(x,y)=2x^3+6xy^2-3y^3-150x ?
The extreme f(x,y)=2x^3+6xy^2-3y^3-150x is Saddle(3,4),Saddle(-3,-4)