Last updated at April 19, 2021 by Teachoo

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 ๐ฅ โ ๐ฅ = 8 โ 2๐ Breadth after removing = 3 โ ๐ฅ โ ๐ฅ = 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) = (8โ2๐ฅ)(3โ2๐ฅ)(๐ฅ) = (8โ2๐ฅ)(3๐ฅโ2๐ฅ2) = 8(3๐ฅโ2๐ฅ2) โ 2x (3๐ฅโ2๐ฅ2) = 24๐ฅ โ 16x2 โ 6๐ฅ2 + 4๐ฅ3 = 4๐3 โ 22๐2 + 24๐ Now, ๐(๐ฅ) = 4๐ฅ3 โ 22๐ฅ2 + 24๐ฅ Diff w.r.t. x ๐โฒ(๐ฅ) = ๐(4๐ฅ^3 โ 22๐ฅ^2 + 24๐ฅ)/๐๐ฅ ๐โฒ(๐ฅ) = 4 ร 3x2 โ 22 ร 2๐ฅ + 24 ๐โฒ(๐ฅ) = 12๐ฅ2 โ 44๐ฅ + 24 ๐โฒ(๐ฅ) = 4(3๐ฅ2โ11๐ฅ+6) Putting ๐ฝโฒ(๐) = 0 4(3๐ฅ2โ11๐ฅ+6) = 0 3๐ฅ2โ11๐ฅ+6 = 0 3๐ฅ2 โ9๐ฅ โ 2๐ฅ + 6 = 0 3๐ฅ(๐ฅโ3) โ2 (๐ฅโ3) = 0 (3๐ฅโ2)(๐ฅโ3)= 0 So, ๐=๐/๐ & ๐=๐ If ๐ = 3 Breadth of a box = 3 โ 2๐ฅ = 3 โ 2(3) = 3 โ 6 = โ3 Since, breadth cannot be negative, โด x = 3 is not possible Hence, ๐ = ๐/๐ only Finding ๐ฝโโ(๐) ๐โ(๐ฅ) = 4(3๐ฅ2โ11๐ฅ+6) Diff w.r.t ๐ฅ ๐โโ(๐ฅ) = ๐(4(3๐ฅ^2 โ 11๐ฅ + 6)/๐๐ฅ ๐โโ(๐ฅ) = 4 (3ร2๐ฅโ11) ๐โโ(๐ฅ) = 4 (6๐ฅโ11) Putting x = ๐/๐ ๐ฝโโ(๐/๐)=4(6(2/3)โ11) = 4 (4โ11)= โ28 < 0 Since ๐โโ(๐ฅ) < 0 at ๐ฅ = 2/3 โด ๐ฅ = 2/3 is point of maxima Hence, ๐ฝ(๐) is largest when ๐ = ๐/๐ Largest volume is ๐(๐ฅ) = x(3โ2๐ฅ) (8โ2๐ฅ) ๐ฝ(๐/๐) = 2/3 (3โ2(2/3)) (8โ2(2/3)) = 2/3 (3โ4/3)(8โ4/3) = 2/3 ((9 โ 4)/3)((24 โ 4)/3) = 2/3 (5/3)(20/3) = 200/27 Since dimension of volume is m3 Largest volume is ๐๐๐/๐๐ m3

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Chapter 6 Class 12 Application of Derivatives

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About the Author

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.