What is the extreme f(x,y)=x^2+2y^2-x+3 ?

The extreme f(x,y)=x^2+2y^2-x+3 is Minimum(1/2 ,0)

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extreme f(x,y)=x^2+2y^2-x+3

extreme points

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

Minimum (\frac{1}{2}, 0)

Solution steps

Find the critical points:

(\frac{1}{2}, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(\frac{1}{2}, 0) :

Minimum

Minimum (\frac{1}{2}, 0)

(x^{2} + 12) (144 - x^{2})

f (x) = \frac{10 x ^{3}}{3} - \frac{11 x ^{2}}{2} + x + 2

5 + 4 x^{2} - x^{4}

f (x) = x^{2} + 3, - 1 \leq x \leq 4

f (x, y) = 5 x + 5 y

What is the extreme f(x,y)=x^2+2y^2-x+3 ?
The extreme f(x,y)=x^2+2y^2-x+3 is Minimum(1/2 ,0)