What is the extreme x^3-6xy+y^2+15x+4 ?

The extreme x^3-6xy+y^2+15x+4 is Minimum(5,15),Saddle(1,3)

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extreme x^3-6xy+y^2+15x+4

extreme points

x^{3} - 6 x y + y^{2} + 15 x + 4

Minimum (5, 15), Saddle (1, 3)

Solution steps

Find the critical points:

(5, 15), (1, 3)

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

D (x, y) = 12 x - 36

Check the sign of at each critical point

Check critical point

(5, 15) :

Minimum

Check critical point

(1, 3) :

Saddle

Minimum (5, 15), Saddle (1, 3)

f (x) = (\frac{1}{12} x^{3} - \frac{1}{2} x^{2} - 12)

50 - 3 x^{2} - 4 x + 2 x y - 8 y^{2} + 78 y

- 3 x^{2} + 12 x + 120

x^{2} + x y + y^{2} - 25 y + 208

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 11

What is the extreme x^3-6xy+y^2+15x+4 ?
The extreme x^3-6xy+y^2+15x+4 is Minimum(5,15),Saddle(1,3)