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Correct option is A)

The curved surface of the cylindrical can be made by rolling the sheet and there will be no weste of material. But in order to have the base of the can, which will be circular of radius r we will have to cut it from a square metal sheet 2r by 2r and whose area will be $4r_{2}$ whereas the actual area of the base is only $πr_{2}$.But no allowance is to be made for waste of material. Hence the total surface area of the sheet used for making the open cylindrical can is $S=2πrh+4r_{2}$

Also $V=πr_{2}h(given)∴h=V/πr_{2}$

$∴S=2πr.πr_{2}V +4r_{2}=r2V +4r_{2}$

$∴drds =−r_{2}2V +8r=0or4r_{3}=V$

$dr_{2}d_{2}S =r_{3}4V +8=4.4+8=24+ive;$

hence minimum when $4r_{3}=V=πr_{2}h∴r/h=π/4$ is the required ratio.

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