What is the extreme f(x,y)=(1-x)(1-y)(x+y-1) ?

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

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extreme f(x,y)=(1-x)(1-y)(x+y-1)

extreme points

f (x, y) = (1 - x) (1 - y) (x + y - 1)

Saddle (1, 0), Maximum (\frac{2}{3}, \frac{2}{3})

Solution steps

Find the critical points:

(1, 0), (\frac{2}{3}, \frac{2}{3})

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

D (x, y) = - 4 x^{2} + 8 x - 4 x y + 8 y - 4 y^{2} - 5

Check the sign of at each critical point

Check critical point

(1, 0) :

Saddle

Check critical point

(\frac{2}{3}, \frac{2}{3}) :

Maximum

Saddle (1, 0), Maximum (\frac{2}{3}, \frac{2}{3})

f (x) = - 6 x^{2} + 5 x y - y^{2} + x + y

x^{4} - 2 x^{2} + 3

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

f (x) = e^{x} (3 - x^{2})

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

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