What is the extreme f(x,y)=e^y-ye^x ?

The extreme f(x,y)=e^y-ye^x is Saddle(0,0)

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extreme f(x,y)=e^y-ye^x

extreme points

f (x, y) = e^{y} - y e^{x}

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = 2 x + cos (2 x), 0 < x < π

f (x, y) = x^{2} - 3 x y + 5 x - 2 y + 6 y^{2} + 8

f (x) = \frac{3}{4} x^{4} - 3 x^{3} - 81 x^{2} - 16

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

f (x) = x^{2} - 5

What is the extreme f(x,y)=e^y-ye^x ?
The extreme f(x,y)=e^y-ye^x is Saddle(0,0)