0.4 is 10 times as much as 0.04 and 0.1 is 10 times as much as 0.01 and 1/10 of 1
0.9 is 10 times as much as 0.09 and 1/10 of 9
sry if my answer is wrong but I think it is the square root of nine
Answer:
(-4,-∞)
Step-by-step explanation:
I think this is what you mean
Answer:
Required largest volume is 0.407114 unit.
Step-by-step explanation:
Given surface area of a right circular cone of radious r and height h is,
and volume,

To find the largest volume if the surface area is S=8 (say), then applying Lagranges multipliers,
subject to,

We know for maximum volume
. So let
be the Lagranges multipliers be such that,



And,



Substitute (3) in (2) we get,



Substitute this value in (1) we get,



Then,

Hence largest volume,
