What is the extreme w(x,y)=xe^2y+e-y ?

The extreme w(x,y)=xe^2y+e-y is Saddle(1/(e^2),0)

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extreme w(x,y)=xe^2y+e-y

extreme points

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

Saddle (\frac{1}{e ^{2}}, 0)

Solution steps

Find the critical points:

(\frac{1}{e ^{2}}, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, y) = - e^{4}

Check the sign of at each critical point

Check critical point

(\frac{1}{e ^{2}}, 0) :

Saddle

Saddle (\frac{1}{e ^{2}}, 0)

f (x) = - 2 x^{2} + 64 x - 80

f (x, y) = x^{2} + y^{2} + 6 x + 10

f (x) = - 34.3 x^{2} + 460.2 x

f (x) = - 3 (5 - 3 x) e^{- 4 x}, (- 1, 4)

f (x) = x^{3} - 6 x^{2} + 18

What is the extreme w(x,y)=xe^2y+e-y ?
The extreme w(x,y)=xe^2y+e-y is Saddle(1/(e^2),0)