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Calculus1 17 Online
OpenStudy (anonymous):

A can in the shape of a cylinder is to have a combined height and circumference of no more than 108 inches to limit shipping costs. Find the dimensions of the can with maximum volum

OpenStudy (anonymous):

is 108 inches an initial volume..?

OpenStudy (anonymous):

maximum volume rather..

OpenStudy (anonymous):

thats what i think, like its 108 dv/dt=3/4pi(r)^3dr/dt, is that right

OpenStudy (anonymous):

u got the wrong formula for the volume of the culinder.. its pi r^2h only..

OpenStudy (anonymous):

oh i was thinking sphere

OpenStudy (anonymous):

so then it would be 108dv/dt=pi(r)^2(h)dr/dt ?

OpenStudy (anonymous):

wait im checking my notes about that problem...

OpenStudy (anonymous):

ok

OpenStudy (fibonaccichick666):

I would just maximize it by looking at the critical points on the derivative, I think.

OpenStudy (anonymous):

its a cylinder, not a graph, and the only thing that looks like a derivative is r^2

OpenStudy (anonymous):

?

OpenStudy (anonymous):

hello

OpenStudy (anonymous):

i think 108 is just a perimeter of the cylinder..

OpenStudy (anonymous):

i mean that is on the problem i think now it might be optimization

OpenStudy (fibonaccichick666):

sorry, ok so max volume, if you take the function for volume and take the derivative then find the critical points and determine where the max would be I think it would work

OpenStudy (anonymous):

can you so step by step really confused with this one

OpenStudy (fibonaccichick666):

Sorry I need sleep but this should help http://mathcentral.uregina.ca/QQ/database/QQ.09.09/h/haris1.html

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