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.

Answer: 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