What is the extreme P(x,y)=x^3(x+y)+5xy(x+y) ?

The extreme P(x,y)=x^3(x+y)+5xy(x+y) is Minimum(3,-12/5),Saddle(5,-5)

extreme P(x,y)=x^3(x+y)+5xy(x+y)

extreme points

P (x, y) = x^{3} (x + y) + 5 x y (x + y)

Minimum (3, - \frac{12}{5}), Saddle (5, - 5)

Solution steps

Find the critical points:

(3, - \frac{12}{5}), (5, - 5)

\frac{\partial ^{2} f}{\partial x ^{2}} = 10 x

D (x, y) = - 9 x^{4} + 60 x^{3} - 100 x^{2} - 100 x y - 100 y^{2}

Check the sign of at each critical point

Check critical point

(3, - \frac{12}{5}) :

Minimum

Check critical point

(5, - 5) :

Saddle

Minimum (3, - \frac{12}{5}), Saddle (5, - 5)

f (x) = - 3 x^{3} + 3 x^{2} - x - 1

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

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

f (x) = sin^{2} (x), 0 \leq x \leq \frac{2 π}{3}

3 x^{2} e^{x}

What is the extreme P(x,y)=x^3(x+y)+5xy(x+y) ?
The extreme P(x,y)=x^3(x+y)+5xy(x+y) is Minimum(3,-12/5),Saddle(5,-5)