What is the extreme f(x,y)=16xy-x^3-8y^2 ?

The extreme f(x,y)=16xy-x^3-8y^2 is Saddle(0,0),Maximum(16/3 , 16/3)

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extreme f(x,y)=16xy-x^3-8y^2

extreme points

f (x, y) = 16 x y - x^{3} - 8 y^{2}

Saddle (0, 0), Maximum (\frac{16}{3}, \frac{16}{3})

Solution steps

Find the critical points:

(0, 0), (\frac{16}{3}, \frac{16}{3})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 16

D (x, y) = 96 x - 256

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{16}{3}, \frac{16}{3}) :

Maximum

Saddle (0, 0), Maximum (\frac{16}{3}, \frac{16}{3})

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

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

f (x) = {4 - 2 x, x \leq 1}

f (x, y) = ln (x^{2} - 4 x + 2 y^{2} - 16 y + 16)

f (x) = - 300 x^{2} + 13500 x - 5400

What is the extreme f(x,y)=16xy-x^3-8y^2 ?
The extreme f(x,y)=16xy-x^3-8y^2 is Saddle(0,0),Maximum(16/3 , 16/3)