you sir are a legend and yes actually funnily enough my modern history teacher recommended this site to me when i had spares today, she said it was epic with calculus etc so obviously she wasn't bsing lol xD thanks man i fully see it now (btw i did have it written down in lined book while i was...
is that the correct answer? I'm sorry i haven't answered your question, i haven't done the limits or the first principles in 2 years :(
also note that the -2 and -3 are meant to be in superscript as powers but i don't know how to work latex like awesomesauce yet
i thought it was because original when it is brought up and r is put to the power of a negative it becomes a negative number...? probably wrong what is the right way to do it i always get mixed up with the order of how the terms and numbers go :(
Homework Statement
Differentiate
Homework Equations
see 1.
The Attempt at a Solution
btw the x in the second equation is meant to be an r
how much did i fail? :(
i don't think this is right or if it is how did you get it so that the second part(0.15(4pir+h) is also multiplied by 0.01 as the 0.01 and 0.015 "segments" of the equation are separated completely by the addition sign? (i'm only first year calculus) so if it's a law or something please explain :D
if Cost(r)=0.01(2\pir2+v/\pir2(2\pir))+0.015(4\pir+v/\pir2)
v=hpir2
how does one show for the optimal radius/height for any volume i.e. where the cost will be the minimum for that volume?
all my attempts have ended with 0=0
help please? i'll give you cookehz ^^
okay so i sort of see what you're saying there compu :)
however we need to find the minimum cost for any volume so yes the optimal height and radius that give the Highest surface area to volume ratio or the smallest values for h and r that give the Vi.e. (the smallest surface area that still...
1. Hey all, For my calculus class we were giving the problem of solving for the optimization of a tin can using differential calculus. The problem was to find the minimum cost for any tin can of any height(as well as using the equation for the tin we had). The surface area of the cylinder was...