Find the dimensions of a right circular cylindrical can with both a top and a bottom that holds 5832 cubic cm and is constructed with the least amount of material possible.

17576=pir^2h the amount of material M=2pir^2 +2pirh M=2pir^2+2pir(17576/pir^2) DrM= 4pir- 35152/r^2 DrM=0 0=4pir- 35152/r^2 r=(13*2^(2/3))/pi^(1/3) 17576=pi((13*2^(2/3))/pi^(1/3))^2)h 17576/(pi((13*2^(2/3))/pi^(1/3))^2))=h r=14.0901 h=28.1801 M=3742.21 cm^2