Example 50 - An open topped box is to be constructed by - Examples

Example 50 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 50 - Chapter 6 Class 12 Application of Derivatives - Part 3 Example 50 - Chapter 6 Class 12 Application of Derivatives - Part 4 Example 50 - Chapter 6 Class 12 Application of Derivatives - Part 5 Example 50 - Chapter 6 Class 12 Application of Derivatives - Part 6

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

Transcript

Example 50 An open topped box is to be constructed by removing equal squares from each corner of a 3 meter by 8 meter rectangular sheet of aluminum and folding up the sides. Find the volume of the largest such box.Let ๐’™ m be the length of a side of the removed square Hence, Length after removing = 8 โ€“ ๐‘ฅ โ€“ ๐‘ฅ = 8 โ€“ 2๐’™ Breadth after removing = 3 โ€“ ๐‘ฅ โ€“ ๐‘ฅ = 3 โ€“ 2๐’™ Height of the box = ๐’™ We need to maximize volume of box Let V be the volume of a box V = Length ร— Breadth ร— Height) = (8โˆ’2๐‘ฅ)(3โˆ’2๐‘ฅ)(๐‘ฅ) = (8โˆ’2๐‘ฅ)(3๐‘ฅโˆ’2๐‘ฅ2) = 8(3๐‘ฅโˆ’2๐‘ฅ2) โ€“ 2x (3๐‘ฅโˆ’2๐‘ฅ2) = 24๐‘ฅ โ€“ 16x2 โ€“ 6๐‘ฅ2 + 4๐‘ฅ3 = 4๐’™3 โ€“ 22๐’™2 + 24๐’™ Now, ๐‘‰(๐‘ฅ) = 4๐‘ฅ3 โ€“ 22๐‘ฅ2 + 24๐‘ฅ Diff w.r.t. x ๐‘‰โ€ฒ(๐‘ฅ) = ๐‘‘(4๐‘ฅ^3 โˆ’ 22๐‘ฅ^2 + 24๐‘ฅ)/๐‘‘๐‘ฅ ๐‘‰โ€ฒ(๐‘ฅ) = 4 ร— 3x2 โ€“ 22 ร— 2๐‘ฅ + 24 ๐‘‰โ€ฒ(๐‘ฅ) = 12๐‘ฅ2 โ€“ 44๐‘ฅ + 24 ๐‘‰โ€ฒ(๐‘ฅ) = 4(3๐‘ฅ2โˆ’11๐‘ฅ+6) Putting ๐‘ฝโ€ฒ(๐’™) = 0 4(3๐‘ฅ2โˆ’11๐‘ฅ+6) = 0 3๐‘ฅ2โˆ’11๐‘ฅ+6 = 0 3๐‘ฅ2 โ€“9๐‘ฅ โ€“ 2๐‘ฅ + 6 = 0 3๐‘ฅ(๐‘ฅโˆ’3) โ€“2 (๐‘ฅโˆ’3) = 0 (3๐‘ฅโˆ’2)(๐‘ฅโˆ’3)= 0 So, ๐’™=๐Ÿ/๐Ÿ‘ & ๐’™=๐Ÿ‘ If ๐’™ = 3 Breadth of a box = 3 โ€“ 2๐‘ฅ = 3 โ€“ 2(3) = 3 โ€“ 6 = โ€“3 Since, breadth cannot be negative, โˆด x = 3 is not possible Hence, ๐’™ = ๐Ÿ/๐Ÿ‘ only Finding ๐‘ฝโ€™โ€™(๐’™) ๐‘‰โ€™(๐‘ฅ) = 4(3๐‘ฅ2โˆ’11๐‘ฅ+6) Diff w.r.t ๐‘ฅ ๐‘‰โ€™โ€™(๐‘ฅ) = ๐‘‘(4(3๐‘ฅ^2 โˆ’ 11๐‘ฅ + 6)/๐‘‘๐‘ฅ ๐‘‰โ€™โ€™(๐‘ฅ) = 4 (3ร—2๐‘ฅโˆ’11) ๐‘‰โ€™โ€™(๐‘ฅ) = 4 (6๐‘ฅโˆ’11) Putting x = ๐Ÿ/๐Ÿ‘ ๐‘ฝโ€™โ€™(๐Ÿ/๐Ÿ‘)=4(6(2/3)โˆ’11) = 4 (4โˆ’11)= โ€“28 < 0 Since ๐‘‰โ€™โ€™(๐‘ฅ) < 0 at ๐‘ฅ = 2/3 โˆด ๐‘ฅ = 2/3 is point of maxima Hence, ๐‘ฝ(๐’™) is largest when ๐’™ = ๐Ÿ/๐Ÿ‘ Largest volume is ๐‘‰(๐‘ฅ) = x(3โˆ’2๐‘ฅ) (8โˆ’2๐‘ฅ) ๐‘ฝ(๐Ÿ/๐Ÿ‘) = 2/3 (3โˆ’2(2/3)) (8โˆ’2(2/3)) = 2/3 (3โˆ’4/3)(8โˆ’4/3) = 2/3 ((9 โˆ’ 4)/3)((24 โˆ’ 4)/3) = 2/3 (5/3)(20/3) = 200/27 Since dimension of volume is m3 Largest volume is ๐Ÿ๐ŸŽ๐ŸŽ/๐Ÿ๐Ÿ• m3

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.