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

The extreme f(x,y)=(x+2y)^y is Saddle(1,0)

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

extreme points

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

Saddle (1, 0)

Solution steps

Find the critical points:

(1, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = (\frac{4 ( x + y + y ^{2} )}{( x + 2 y ) ^{2}} + \frac{4 y ln ( x + 2 y )}{x + 2 y} + ln^{2} (x + 2 y)) (x + 2 y)^{y}

D (x, y) = - 8 y^{4} ln (x + 2 y) (x + 2 y)^{2 y - 4} - 4 y^{4} ln^{2} (x + 2 y) (x + 2 y)^{2 y - 4} - 4 x y^{3} ln (x + 2 y) (x + 2 y)^{2 y - 4} - 4 x y^{3} ln^{2} (x + 2 y) (x + 2 y)^{2 y - 4} + 4 y^{3} ln (x + 2 y) (x + 2 y)^{2 y - 3} - x^{2} y^{2} ln^{2} (x + 2 y) (x + 2 y)^{2 y - 4} - 4 x y^{2} ln (x + 2 y) (x + 2 y)^{2 y - 4} - 4 y^{2} ln (x + 2 y) (x + 2 y)^{2 y - 3} + y^{2} ln^{2} (x + 2 y) (x + 2 y)^{2 y - 2} - 2 x^{2} y ln (x + 2 y) (x + 2 y)^{2 y - 4} - y ln^{2} (x + 2 y) (x + 2 y)^{2 y - 2} - (x + 2 y)^{2 (y - 1)}

Check the sign of at each critical point

Check critical point

(1, 0) :

Saddle

Saddle (1, 0)

f (x) = x^{4} - 16 x^{3}

f (x) = \frac{2 ^{2}}{2 ^{4} + 16}

f (x, y) = 2 x^{2} - x y - 3 y^{2} - 3 x + 7 y

f (x, y) = x^{3} + 6 x y + y^{3}

f (x, y) = x y - 5 x + 15

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