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CBSE Class 12-science Answered

Suppose a certain sheet of metal is given and we need to make a right circular cylinder tin out of it such that it contains maximum quantity of oil. Here we assume no metal is wasted.
Asked by Topperlearning User | 19 Aug, 2014, 09:12: AM
answered-by-expert Expert Answer

Assuming space that space the space required space cylinder space of space radius space straight r space and space height space straight h. space Since space the space sheet of space metal space is space given space so space we space can space say space that space the space total space surface space area space is space given space therefore straight S equals 2 πrh plus 2 πr squared equals constant. space... left parenthesis straight i right parenthesis Let space straight V space be space the space volume space of space the space cylinder comma space which space we space need space to space maximise straight V equals πr squared straight h This space equation space has space two space variable space straight r space and space straight h. straight h equals fraction numerator straight S minus 2 πr squared over denominator 2 πr end fraction space left square bracket From space left parenthesis straight i right parenthesis right square bracket therefore straight V equals πr squared open parentheses fraction numerator straight S minus 2 πr squared over denominator 2 πr end fraction close parentheses equals fraction numerator Sr minus 2 πr squared over denominator 2 end fraction equals Sr over 2 minus πr cubed We space need space to space maximise space the space volume therefore dV over dr equals straight S over 2 minus 3 πr squared rightwards double arrow dV over dr equals 0 rightwards double arrow straight S over 2 minus 3 πr squared equals 0 rightwards double arrow straight S equals 6 πr squared Substituting space straight S comma space we space get 2 πrh plus 2 πr squared equals 6 πr squared rightwards double arrow 2 πrh equals 4 πr squared rightwards double arrow straight h equals 2 straight r To space check space for space maxima space or space minima comma space fraction numerator straight d squared straight V over denominator dr squared end fraction equals minus 6 πr comma space which space is space negative space as space straight r space is positive comma space so space capacity space of space cylinder space will space be space maximum space when space diameter space of space base space equals height space of space cylinder.

Answered by | 19 Aug, 2014, 11:12: AM

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