What is the extreme f(x,y)=4x^2+8y^2 ?

The extreme f(x,y)=4x^2+8y^2 is Minimum(0,0)

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extreme f(x,y)=4x^2+8y^2

extreme points

f (x, y) = 4 x^{2} + 8 y^{2}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 16

D (x, y) = 128

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = \frac{2 x + 1 - 18}{x}

f (x, y) = - 11 \cdot y^{2} + (x + 16)^{2} + 1

\frac{x ^{3}}{( x - 1 ) ^{2}}

x^{3} - 6 x^{2} + 8

f (x) = x^{2} - 8 x + 7

What is the extreme f(x,y)=4x^2+8y^2 ?
The extreme f(x,y)=4x^2+8y^2 is Minimum(0,0)