What is the extreme y=2x^2+z^2 ?

The extreme y=2x^2+z^2 is Minimum(0,0)

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extreme y=2x^2+z^2

extreme points

y = 2 x^{2} + z^{2}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, z) = 8

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x, y) = 1 - 3 x^{4} - 6 x^{2} y + 2 y^{3}

f (x) = 8 x^{3} + 13 x

f (x, y) = 18 y^{2} + x^{2} - x^{2} y

2 (\frac{1}{2})^{x}

f (x) = \frac{( x - 4 ) ^{2}}{x + 10}