What is the extreme f(x)=x^3-48xy+64y^3 ?

The extreme f(x)=x^3-48xy+64y^3 is Saddle(0,0),Minimum(4,1)

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extreme f(x)=x^3-48xy+64y^3

extreme points

f (x) = x^{3} - 48 x y + 64 y^{3}

Saddle (0, 0), Minimum (4, 1)

Solution steps

Find the critical points:

(0, 0), (4, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 384 y

D (x, y) = 2304 x y - 2304

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(4, 1) :

Minimum

Saddle (0, 0), Minimum (4, 1)

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

e^{x} (x - 3)

y = 2 sin (x) + 9 cos (x), 0 \leq x \leq 2 π

f (x) = - 10 x^{2} + 180 x + 10

f (x) = \frac{8 - x}{5 + x}

What is the extreme f(x)=x^3-48xy+64y^3 ?
The extreme f(x)=x^3-48xy+64y^3 is Saddle(0,0),Minimum(4,1)