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

The extreme f(x,y)=x^3-4xy+y^2 is Minimum(8/3 , 16/3),Saddle(0,0)

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

extreme points

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

Minimum (\frac{8}{3}, \frac{16}{3}), Saddle (0, 0)

Solution steps

Find the critical points:

(\frac{8}{3}, \frac{16}{3}), (0, 0)

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

D (x, y) = 12 x - 16

Check the sign of at each critical point

Check critical point

(\frac{8}{3}, \frac{16}{3}) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (\frac{8}{3}, \frac{16}{3}), Saddle (0, 0)

f (t) = - 4 t^{2} + 208 t + 120

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

f (x) = x (\frac{x ^{2}}{5} - 4)

- x^{2} - 6 x - 4

f (x) = \frac{x}{x - 9}, 11 \leq x \leq 13

What is the extreme f(x,y)=x^3-4xy+y^2 ?
The extreme f(x,y)=x^3-4xy+y^2 is Minimum(8/3 , 16/3),Saddle(0,0)