What is the extreme f(y)=x^3+y^3-27xy ?

The extreme f(y)=x^3+y^3-27xy is Saddle(0,0),Minimum(9,9)

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extreme f(y)=x^3+y^3-27xy

extreme points

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

Saddle (0, 0), Minimum (9, 9)

Solution steps

Find the critical points:

(0, 0), (9, 9)

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

D (y, x) = 36 y x - 729

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(9, 9) :

Minimum

Saddle (0, 0), Minimum (9, 9)

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

f (x) = 12 x^{2} - 88 x + 121

f (x) = 9 x^{3} - 7 x^{2} + 3 x + 10, - 5 \leq x \leq 6

f (x, y) = 400 - 49 x^{2} - 9 y^{2}

f (x) = 2 x^{3} + 24 x^{2} + 72 x + 7

What is the extreme f(y)=x^3+y^3-27xy ?
The extreme f(y)=x^3+y^3-27xy is Saddle(0,0),Minimum(9,9)