What is the extreme f(x,y)=8xy-x^3-4y^2 ?

The extreme f(x,y)=8xy-x^3-4y^2 is Saddle(0,0),Maximum(8/3 , 8/3)

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extreme f(x,y)=8xy-x^3-4y^2

extreme points

f (x, y) = 8 x y - x^{3} - 4 y^{2}

Saddle (0, 0), Maximum (\frac{8}{3}, \frac{8}{3})

Solution steps

Find the critical points:

(0, 0), (\frac{8}{3}, \frac{8}{3})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 8

D (x, y) = 48 x - 64

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{8}{3}, \frac{8}{3}) :

Maximum

Saddle (0, 0), Maximum (\frac{8}{3}, \frac{8}{3})

f (x) = 5 x^{6} - 3 x^{4} + 2 x - 9

f (x, y) = x^{3} + y^{3} - 9 x y + 27

f (x) = 6 x^{2} ln (x)

x^{4} - 4 x^{3} + 1

f (x, y) = \frac{x ^{4} - y ^{4}}{x ^{2} - y ^{2}}

What is the extreme f(x,y)=8xy-x^3-4y^2 ?
The extreme f(x,y)=8xy-x^3-4y^2 is Saddle(0,0),Maximum(8/3 , 8/3)