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

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

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

extreme points

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

Minimum (\frac{1}{2}, - \frac{1}{4})

Solution steps

Find the critical points:

(\frac{1}{2}, - \frac{1}{4})

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

D (x, y) = 16

Check the sign of at each critical point

Check critical point

(\frac{1}{2}, - \frac{1}{4}) :

Minimum

Minimum (\frac{1}{2}, - \frac{1}{4})

f (x) = - \frac{7}{x ^{2} + 1}

f (x) = \frac{x}{25 - x + x ^{2}}

f (x) = (x^{2} - 25)^{2}

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

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

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