What is the extreme f(x,y)=12xy-x^3-6y^2 ?

The extreme f(x,y)=12xy-x^3-6y^2 is Saddle(0,0),Maximum(4,4)

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extreme f(x,y)=12xy-x^3-6y^2

extreme points

f (x, y) = 12 x y - x^{3} - 6 y^{2}

Saddle (0, 0), Maximum (4, 4)

Solution steps

Find the critical points:

(0, 0), (4, 4)

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

D (x, y) = 72 x - 144

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(4, 4) :

Maximum

Saddle (0, 0), Maximum (4, 4)

x^{3} - x^{2}

f (x, y) = x y + \frac{27}{x} + \frac{27}{y}

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

f (x) = 8 x - x^{4}

f (x) = x^{2} - 2 x - 4

What is the extreme f(x,y)=12xy-x^3-6y^2 ?
The extreme f(x,y)=12xy-x^3-6y^2 is Saddle(0,0),Maximum(4,4)