What is the extreme f(x,y)=xey^2+yln(x) ?

The extreme f(x,y)=xey^2+yln(x) is Minimum(e^2,-1/(e^3)),Saddle(1,0)

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extreme f(x,y)=xey^2+yln(x)

extreme points

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

Minimum (e^{2}, - \frac{1}{e ^{3}}), Saddle (1, 0)

Solution steps

Find the critical points:

(e^{2}, - \frac{1}{e ^{3}}), (1, 0)

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

D (x, y) = \frac{- 4 e ^{2} x ^{2} y ^{2} - 6 e x y - 1}{x ^{2}}

Check the sign of at each critical point

Check critical point

(e^{2}, - \frac{1}{e ^{3}}) :

Minimum

Check critical point

(1, 0) :

Saddle

Minimum (e^{2}, - \frac{1}{e ^{3}}), Saddle (1, 0)

y = - x^{2} - x - 5

f (x) = 3 x^{4} - 8 x^{3} + 6 x^{2} - 9

f (x, y) = 7 x^{2} y + 9 x y^{2}

- 4 x^{3} + 200 x^{2} - 3000 x + 17000

f (x) = - x + 10

What is the extreme f(x,y)=xey^2+yln(x) ?
The extreme f(x,y)=xey^2+yln(x) is Minimum(e^2,-1/(e^3)),Saddle(1,0)