What is the extreme f(xy)=2x^2-2xy+y^2+5x-3y ?

The extreme f(xy)=2x^2-2xy+y^2+5x-3y is Minimum(-1, 1/2)

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extreme f(xy)=2x^2-2xy+y^2+5x-3y

extreme points

f (x y) = 2 x^{2} - 2 x y + y^{2} + 5 x - 3 y

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

Solution steps

Find the critical points:

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

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

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

Minimum

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

f (x) = (1 - e^{- \frac{x \cdot 3}{5}})^{x}

f (x) = 5 x^{2} + 4 x - 1

- x - 2

y = x^{6} (x - 6)^{7}

5.11 x - x x - 2

What is the extreme f(xy)=2x^2-2xy+y^2+5x-3y ?
The extreme f(xy)=2x^2-2xy+y^2+5x-3y is Minimum(-1, 1/2)