What is the extreme f(x,y)=x^2y+2y^2-2xy+6 ?

The extreme f(x,y)=x^2y+2y^2-2xy+6 is Saddle(2,0),Saddle(0,0),Minimum(1, 1/4)

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extreme f(x,y)=x^2y+2y^2-2xy+6

extreme points

f (x, y) = x^{2} y + 2 y^{2} - 2 x y + 6

Saddle (2, 0), Saddle (0, 0), Minimum (1, \frac{1}{4})

Solution steps

Find the critical points:

(2, 0), (0, 0), (1, \frac{1}{4})

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

D (x, y) = 8 y - 4 x^{2} + 8 x - 4

Check the sign of at each critical point

Check critical point

(2, 0) :

Saddle

Check critical point

(0, 0) :

Saddle

Check critical point

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

Minimum

Saddle (2, 0), Saddle (0, 0), Minimum (1, \frac{1}{4})

f (x) = x (10 + x), - 1 \leq x \leq 3

y = x (10 - 2 x) (8 - x)

f (x) = \frac{3 x ^{2} - 16 x + 16}{( x ^{2} - 2 x ) ^{2}}

- x^{\frac{2}{3}} (x - 5)

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

What is the extreme f(x,y)=x^2y+2y^2-2xy+6 ?
The extreme f(x,y)=x^2y+2y^2-2xy+6 is Saddle(2,0),Saddle(0,0),Minimum(1, 1/4)