What is the extreme x^3+y^2+12xy+1 ?

The extreme x^3+y^2+12xy+1 is Saddle(0,0),Minimum(24,-144)

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extreme x^3+y^2+12xy+1

extreme points

x^{3} + y^{2} + 12 x y + 1

Saddle (0, 0), Minimum (24, - 144)

Solution steps

Find the critical points:

(0, 0), (24, - 144)

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

D (x, y) = 12 x - 144

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(24, - 144) :

Minimum

Saddle (0, 0), Minimum (24, - 144)

f (x) = 200 x - (0.003 x^{2} + 80 x + 500000)

- \frac{\frac{40}{3}}{x ^{2}} + 0.54 x

f (x, y) = x \cdot e^{y}

f (x) = 4 θ - 6 sin (θ)

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

What is the extreme x^3+y^2+12xy+1 ?
The extreme x^3+y^2+12xy+1 is Saddle(0,0),Minimum(24,-144)