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

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

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extreme x^2(2+y^2)+yln(y)

extreme points

x^{2} (2 + y^{2}) + y ln (y)

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})

3 x 2 + 3 y 2 + 3 z 2 + 2 y - 2 z = 9

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

f (x) = 0.00324 x^{2} - 0.501 x + 33.684

f (x, y) = (x - y)^{2} - x^{4} - y

f (x) = x^{3} - x^{2} - x + 9, - 1 \leq x \leq 2

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