What is the extreme f(x,y)=2x^2+7y^2 ?

The extreme f(x,y)=2x^2+7y^2 is Minimum(0,0)

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extreme f(x,y)=2x^2+7y^2

extreme points

f (x, y) = 2 x^{2} + 7 y^{2}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 14

D (x, y) = 56

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

5 (5 + x)^{4}

f (x) = 2 - 2 x^{2}, - 4 \leq x \leq 1

f (x) = 3 x + sin (3 x)

f (x) = 6 x^{3} - 9 x^{2} - 216 x + 7, - 4 \leq x \leq 5

e^{x^{2} + 3 \cdot x + 4}

What is the extreme f(x,y)=2x^2+7y^2 ?
The extreme f(x,y)=2x^2+7y^2 is Minimum(0,0)