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

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

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

extreme

Minimum (0, \frac{1}{e})

Solution steps

Find the critical points:

(0, \frac{1}{e})

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{1}{y} + 2 x^{2}

D (x, y) = - 12 x^{2} y^{2} + 2 y + \frac{4}{y} + 8 x^{2}

Check the sign of at each critical point

Check critical point

(0, \frac{1}{e}) :

Minimum

Minimum (0, \frac{1}{e})

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