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

The extreme f(x,y)=xe-2x^2-2y^2 is Maximum(e/4 ,0)

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extreme f(x,y)=xe-2x^2-2y^2

extreme points

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

Maximum (\frac{e}{4}, 0)

Solution steps

Find the critical points:

(\frac{e}{4}, 0)

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

D (x, y) = 16

Check the sign of at each critical point

Check critical point

(\frac{e}{4}, 0) :

Maximum

Maximum (\frac{e}{4}, 0)

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

f (t) = t u (t - 2)

f (x) = x^{3} - 5 x^{2} - 8 x

f (x, y) = - 2 y - 3 x + 6

f (x) = \frac{( x - 2 ) ^{2}}{x + 6}

What is the extreme f(x,y)=xe-2x^2-2y^2 ?
The extreme f(x,y)=xe-2x^2-2y^2 is Maximum(e/4 ,0)