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

The extreme f(x)=8x^3-3xy+y^3 is Saddle(0,0),Minimum(1/4 , 1/2)

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extreme f(x)=8x^3-3xy+y^3

extreme points

f (x) = 8 x^{3} - 3 x y + y^{3}

Saddle (0, 0), Minimum (\frac{1}{4}, \frac{1}{2})

Solution steps

Find the critical points:

(0, 0), (\frac{1}{4}, \frac{1}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y

D (x, y) = 288 x y - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{1}{4}, \frac{1}{2}) :

Minimum

Saddle (0, 0), Minimum (\frac{1}{4}, \frac{1}{2})

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

y (x, t) = 7 x - 8 a t (- 4.5)

- x^{2} + 5 x - 2

y = 9 - x

f (x, y) = 49 - x^{2} - y^{2}

What is the extreme f(x)=8x^3-3xy+y^3 ?
The extreme f(x)=8x^3-3xy+y^3 is Saddle(0,0),Minimum(1/4 , 1/2)