What is the extreme f(x,y)=2x^2+4y^2+1/2 ?

The extreme f(x,y)=2x^2+4y^2+1/2 is Minimum(0,0)

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extreme f(x,y)=2x^2+4y^2+1/2

extreme points

f (x, y) = 2 x^{2} + 4 y^{2} + \frac{1}{2}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 8

D (x, y) = 32

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = \frac{x ^{2} + 2}{x + 1}

f (x) = 2 x - 11 ln (8 x), 2 \leq x \leq 9

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

(2 x - x^{2}) (2 y - y^{2})

sin (2 x)

What is the extreme f(x,y)=2x^2+4y^2+1/2 ?
The extreme f(x,y)=2x^2+4y^2+1/2 is Minimum(0,0)