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

The extreme f(x)=f(x,y)=xye^{x+2y} is Maximum(-1,-1/2)

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

extreme points

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

Maximum (- 1, - \frac{1}{2})

Solution steps

Find the critical points:

(- 1, - \frac{1}{2})

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

D (x, y) = x y (e^{x + 2 y} x + 2 e^{x + 2 y}) (4 e^{x + 2 y} y + 4 e^{x + 2 y}) - e^{2 (2 y + x)} (1 + x)^{2} (1 + 2 y)^{2}

Check the sign of at each critical point

Check critical point

(- 1, - \frac{1}{2}) :

Maximum

Maximum (- 1, - \frac{1}{2})

y = x 64 - x^{2}

f (x) = \frac{1}{3} x^{3} + \frac{5}{2} x^{2} - 24 x + 2

f (x) = - 3 x^{2} + 3 x y - 2 x + y^{2} + 6 y + 12

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

f (x) = 4 - x

What is the extreme f(x)=f(x,y)=xye^{x+2y} ?
The extreme f(x)=f(x,y)=xye^{x+2y} is Maximum(-1,-1/2)