What is the extreme f(s,t)=s^3-2t^2+8st ?

The extreme f(s,t)=s^3-2t^2+8st is Saddle(0,0),Maximum(-16/3 ,-32/3)

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extreme f(s,t)=s^3-2t^2+8st

extreme points

f (s, t) = s^{3} - 2 t^{2} + 8 s t

Saddle (0, 0), Maximum (- \frac{16}{3}, - \frac{32}{3})

Solution steps

Find the critical points:

(0, 0), (- \frac{16}{3}, - \frac{32}{3})

\frac{\partial ^{2} f}{\partial s ^{2}} = - 4

D (s, t) = - 24 s - 64

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- \frac{16}{3}, - \frac{32}{3}) :

Maximum

Saddle (0, 0), Maximum (- \frac{16}{3}, - \frac{32}{3})

f (z) = \frac{1}{2} y^{4} - 4 x y + 2 x^{4}

f (x, y) = x y + y

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

f (x) = x \cdot 4 - x

f (x) = x (x - 5)^{2} (x + 1)

What is the extreme f(s,t)=s^3-2t^2+8st ?
The extreme f(s,t)=s^3-2t^2+8st is Saddle(0,0),Maximum(-16/3 ,-32/3)