What is the extreme f(x)=x^2y+y^3-75y ?

The extreme f(x)=x^2y+y^3-75y is Minimum(0,5),Maximum(0,-5),Saddle(5sqrt(3),0),Saddle(-5sqrt(3),0)

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extreme f(x)=x^2y+y^3-75y

extreme points

f (x) = x^{2} y + y^{3} - 75 y

Minimum (0, 5), Maximum (0, - 5), Saddle (53, 0), Saddle (- 53, 0)

Solution steps

Find the critical points:

(0, 5), (0, - 5), (53, 0), (- 53, 0)

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

D (x, y) = 12 y^{2} - 4 x^{2}

Check the sign of at each critical point

Check critical point

(0, 5) :

Minimum

Check critical point

(0, - 5) :

Maximum

Check critical point

(53, 0) :

Saddle

Check critical point

(- 53, 0) :

Saddle

Minimum (0, 5), Maximum (0, - 5), Saddle (53, 0), Saddle (- 53, 0)

f (x, y) = x^{2} + x y + y^{2} - 4 y + 5

y = x^{3} - 3 x^{2} + 1

f (x, y) = x^{3} + y^{3} - 33 x y

4 x y - x^{4} + y^{4}

f (x) = - 2 x^{2} - 16 x - 39

What is the extreme f(x)=x^2y+y^3-75y ?
The extreme f(x)=x^2y+y^3-75y is Minimum(0,5),Maximum(0,-5),Saddle(5sqrt(3),0),Saddle(-5sqrt(3),0)