What is the extreme f(x,y)=4y^3x^2-12y^2-36y+2 ?

The extreme f(x,y)=4y^3x^2-12y^2-36y+2 is Maximum(0,-3/2)

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extreme f(x,y)=4y^3x^2-12y^2-36y+2

extreme

Maximum (0, - \frac{3}{2})

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = 24 x^{2} y - 24

D (x, y) = - 384 x^{2} y^{4} - 192 y^{3}

Check the sign of at each critical point

Check critical point

(0, - \frac{3}{2}) :

Maximum

Maximum (0, - \frac{3}{2})

What is the extreme f(x,y)=4y^3x^2-12y^2-36y+2 ?
The extreme f(x,y)=4y^3x^2-12y^2-36y+2 is Maximum(0,-3/2)