What is the extreme f(x,y)=e^y(4x^2+2xy+y^2-6x-15) ?

The extreme f(x,y)=e^y(4x^2+2xy+y^2-6x-15) is Minimum(0,3),Saddle(5/2 ,-7)

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extreme f(x,y)=e^y(4x^2+2xy+y^2-6x-15)

extreme

Minimum (0, 3), Saddle (\frac{5}{2}, - 7)

Solution steps

Find the critical points:

(0, 3), (\frac{5}{2}, - 7)

\frac{\partial ^{2} f}{\partial x ^{2}} = e^{y} (4 x^{2} + 2 x y - 2 x + y^{2} + 4 y - 13)

D (x, y) = - 32 e^{2 y} x^{2} + 48 e^{2 y} x - 16 e^{2 y} x y - 120 e^{2 y} + 4 e^{2 y} y^{2} + 48 e^{2 y} y

Check the sign of at each critical point

Check critical point

(0, 3) :

Minimum

Check critical point

(\frac{5}{2}, - 7) :

Saddle

Minimum (0, 3), Saddle (\frac{5}{2}, - 7)

What is the extreme f(x,y)=e^y(4x^2+2xy+y^2-6x-15) ?
The extreme f(x,y)=e^y(4x^2+2xy+y^2-6x-15) is Minimum(0,3),Saddle(5/2 ,-7)