What is the extreme f(x,y)=x^2y+y^3-48y ?

The extreme f(x,y)=x^2y+y^3-48y is Minimum(0,4),Maximum(0,-4),Saddle(4sqrt(3),0),Saddle(-4sqrt(3),0)

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extreme f(x,y)=x^2y+y^3-48y

extreme points

f (x, y) = x^{2} y + y^{3} - 48 y

Minimum (0, 4), Maximum (0, - 4), Saddle (43, 0), Saddle (- 43, 0)

Solution steps

Find the critical points:

(0, 4), (0, - 4), (43, 0), (- 43, 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, 4) :

Minimum

Check critical point

(0, - 4) :

Maximum

Check critical point

(43, 0) :

Saddle

Check critical point

(- 43, 0) :

Saddle

Minimum (0, 4), Maximum (0, - 4), Saddle (43, 0), Saddle (- 43, 0)

f (x) = - 5 x^{4} - 5 x^{3} + 29

f (x) = 7 sin (x) + 7 cos (x), 0 \leq x \leq 2 π

x^{3} - 6 x^{2} + 1

f (x) = 2 x^{3} + 12 x^{2} + 18 x

f (x) = \frac{4}{3} x^{3} + 7 x^{2} - 8 x - 42

What is the extreme f(x,y)=x^2y+y^3-48y ?
The extreme f(x,y)=x^2y+y^3-48y is Minimum(0,4),Maximum(0,-4),Saddle(4sqrt(3),0),Saddle(-4sqrt(3),0)