What is the extreme f(r,h)=2pirh+2pir^2 ?

The extreme f(r,h)=2pirh+2pir^2 is Saddle(0,0)

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extreme f(r,h)=2pirh+2pir^2

extreme points

f (r, h) = 2 π r h + 2 π r^{2}

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (r, h) = - 4 π^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

g (t) = u (t)

f (x) = x^{3} + 10 x - 7

f (x) = 9 x - 18 cos (x)

x^{1} - x^{2} + 2 \cdot x 1^{2} + 2 \cdot x^{1} \cdot x^{2} + x 2^{2}

What is the extreme f(r,h)=2pirh+2pir^2 ?
The extreme f(r,h)=2pirh+2pir^2 is Saddle(0,0)