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

The extreme f(x,y)=(x^2+y)e^{y/2} is Minimum(0,-2)

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

extreme points

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

Minimum (0, - 2)

Solution steps

Find the critical points:

(0, - 2)

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

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

Check the sign of at each critical point

Check critical point

(0, - 2) :

Minimum

Minimum (0, - 2)

h (t, v) = 5 t^{2} + v t

x 5 - x

- x^{3} + 6 x^{2} - 9 x

ln (u)

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

What is the extreme f(x,y)=(x^2+y)e^{y/2} ?
The extreme f(x,y)=(x^2+y)e^{y/2} is Minimum(0,-2)