What is the extreme f(x,y)=3xe^y-x^3-e^{3y} ?

The extreme f(x,y)=3xe^y-x^3-e^{3y} is Maximum(1,0)

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extreme f(x,y)=3xe^y-x^3-e^{3y}

extreme points

f (x, y) = 3 x e^{y} - x^{3} - e^{3 y}

Maximum (1, 0)

Solution steps

Find the critical points:

(1, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 3 x e^{y} - 9 e^{3 y}

D (x, y) = - 6 x (3 e^{y} x - 9 e^{3 y}) - 9 e^{2 y}

Check the sign of at each critical point

Check critical point

(1, 0) :

Maximum

Maximum (1, 0)

f (x, y) = x^{2} - 4 x y + y^{3} + 4 y

f (x) = x^{3} + 2 x^{2} - 15 x - 20

f (x, y) = x^{3} + 3 x y^{2} - 15 x - 12 y

f (x, y) = ln (x) + 2 ln (y) - x - 4 y

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

What is the extreme f(x,y)=3xe^y-x^3-e^{3y} ?
The extreme f(x,y)=3xe^y-x^3-e^{3y} is Maximum(1,0)