解析:如图,矩形ABCD是过圆柱的高的截面,设∠AOD=2θ,则AD=2Rsinθ,AB=2Rcosθ,
∴S圆柱全=2·π()2+π·AD·AB=2πR2sin2θ+π·4R2sinθcosθ=πR2(1-cos2θ+2sin2θ)+πR2sin(2θ-arctan)+πR2.
∴当2θ-arctan=,即θ=+·arctan时,S圆柱全=πR2(+1)为最大值.