What is the extreme f(x,y)=2x^3y-29x^2-16y ?

The extreme f(x,y)=2x^3y-29x^2-16y is Saddle(2, 29/6)

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extreme f(x,y)=2x^3y-29x^2-16y

extreme points

f (x, y) = 2 x^{3} y - 29 x^{2} - 16 y

Saddle (2, \frac{29}{6})

Solution steps

Find the critical points:

(2, \frac{29}{6})

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, y) = - 36 x^{4}

Check the sign of at each critical point

Check critical point

(2, \frac{29}{6}) :

Saddle

Saddle (2, \frac{29}{6})

y = x^{2} + 2 x + 25

y = I n (x)

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

cos^{2} (x) + sin (x)

f (x) = 2 x - 1

What is the extreme f(x,y)=2x^3y-29x^2-16y ?
The extreme f(x,y)=2x^3y-29x^2-16y is Saddle(2, 29/6)