What is the extreme f(x)=x^3+y^3-15xy ?

The extreme f(x)=x^3+y^3-15xy is Saddle(0,0),Minimum(5,5)

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extreme f(x)=x^3+y^3-15xy

extreme points

f (x) = x^{3} + y^{3} - 15 x y

Saddle (0, 0), Minimum (5, 5)

Solution steps

Find the critical points:

(0, 0), (5, 5)

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

D (x, y) = 36 x y - 225

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(5, 5) :

Minimum

Saddle (0, 0), Minimum (5, 5)

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

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

f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} + 6 x - 7 y

y = x e^{\frac{x}{2}}

f (x) = x^{4} - 62 x^{2} + 120 x + 9

What is the extreme f(x)=x^3+y^3-15xy ?
The extreme f(x)=x^3+y^3-15xy is Saddle(0,0),Minimum(5,5)