What is the extreme f(x,y)=25x^2+4y^2+4 ?

The extreme f(x,y)=25x^2+4y^2+4 is Minimum(0,0)

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extreme f(x,y)=25x^2+4y^2+4

extreme points

f (x, y) = 25 x^{2} + 4 y^{2} + 4

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 400

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = - 4 x^{3} + 12 x^{2} + 4 x - 12

f (x) = \frac{4 x - 3}{x ^{2} - 4 x + 4}

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 4

S (a, d) = \frac{n}{2} [2 a + (n - 1) d]

f (x) = x^{3} - 12 x^{2} + 36 x - 1

What is the extreme f(x,y)=25x^2+4y^2+4 ?
The extreme f(x,y)=25x^2+4y^2+4 is Minimum(0,0)