can't solve this equation

Sentinel

Diamond Member
Jun 23, 2000
3,714
1
71
err...derivatives.

4(pi)r - 32(pi)r^-2 = 0

all i need to find is ie. (x + 1) (x - 1)

pi ~ 3.14

trying to find the critical numbers so I can get max and min in f''(x)
 

Sentinel

Diamond Member
Jun 23, 2000
3,714
1
71
the question is as follows:

a large soup can is to be designed so that the can will hold 16(pi) cubic inches of soup. Find the values of x and h such that the amount of metal needed is as small as possible.

the side requieres 2(pi)xh square inches of metal and each end requires (pi)x^2 square inches.

 

duragezic

Lifer
Oct 11, 1999
11,234
4
81
Originally posted by: Sentinel
nice :thumbsup:

f'(x) = 4(pi)r(-8r^-1 + 1)
What? Multiply both terms on the left from the original equation by r^2 and you get 4pir^3 for the first and the r^-2 cancels leaving you just -32pi as the 2nd term. Right-side is zero anyway.
 

hypn0tik

Diamond Member
Jul 5, 2005
5,866
2
0
Originally posted by: Heisenberg
Originally posted by: ActuaryTm
Too difficult. Math is hard.
Yeah really. All those numbers and variables and stuff. It's really stupid.

This coming from a guy whose username is 'Heisenberg'. Lol.
 

ActuaryTm

Diamond Member
Mar 30, 2003
6,858
12
81
Originally posted by: hypn0tik
This coming from a guy whose username is 'Heisenberg'. Lol.
Try looking up the word "actuary".

Heisenberg's sarcasm may become a bit more apparent.