Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12


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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise


Ex 6.5, 18 A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum ? Let ๐‘ฅ be the length of a side of the removed square Thus, Length after removing = 45 โ€“ ๐‘ฅ โ€“๐‘ฅ = 45 โ€“ 2๐‘ฅ Breadth after removing = 24 โ€“๐‘ฅ โ€“๐‘ฅ = 24 โ€“ 2๐‘ฅ Height of the box = ๐‘ฅ We need to maximize volume of box Let V be the volume of a box Volume of a cuboid = l ร— b ร— h = (45โˆ’2๐‘ฅ)(24โˆ’2๐‘ฅ)(๐‘ฅ) = [45(24โˆ’2๐‘ฅ)โˆ’2๐‘ฅ(24โˆ’2๐‘ฅ)]๐‘ฅ = (1080โˆ’90๐‘ฅโˆ’48๐‘ฅ+4๐‘ฅ^2 )๐‘ฅ = 1080๐‘ฅโˆ’90๐‘ฅ^2โˆ’48๐‘ฅ^2+4๐‘ฅ^3 = 1080๐‘ฅโˆ’138๐‘ฅ^2+4๐‘ฅ^3 = 2(540๐‘ฅโˆ’69๐‘ฅ^2+2๐‘ฅ^3 ) = 2(2๐‘ฅ^3โˆ’69๐‘ฅ^2+540๐‘ฅ) V = 2(2๐‘ฅ^3โˆ’69๐‘ฅ^2+540) Diff w.r.t ๐‘ฅ Vโ€™(๐‘ฅ)=2๐‘‘[2๐‘ฅ^3โˆ’ 69๐‘ฅ^2+ 540๐‘ฅ]/๐‘‘๐‘ฅ = 2[6๐‘ฅ^2โˆ’69 ร—2๐‘ฅ+540] = 2[6๐‘ฅ^2โˆ’138๐‘ฅ+540] = 2 ร— 6[๐‘ฅ^2โˆ’23๐‘ฅ+90] = 12[๐‘ฅ^2โˆ’23๐‘ฅ+90] Putting Vโ€™(๐‘ฅ)=0 12(๐‘ฅ^2โˆ’23๐‘ฅ+90)=0 ๐‘ฅ^2โˆ’23๐‘ฅ+90=0 ๐‘ฅ^2โˆ’5๐‘ฅโˆ’18๐‘ฅ+90=0 ๐‘ฅ(๐‘ฅโˆ’5)โˆ’18(๐‘ฅโˆ’5)=0 (๐‘ฅโˆ’18)(๐‘ฅโˆ’5)=0 So, x = 18 & x = 5 Thus, V(๐‘ฅ) is maximum at ๐‘ฅ=5 โˆด Square of side 5 cm is cut off from each Corner

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.