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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

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

Minimum

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

f (x, y) = \frac{1}{2} x^{2} - x y + \frac{1}{3} y^{3}

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

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

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

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

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