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Last updated at Jan. 7, 2020 by Teachoo

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Ex 6.5, 17 A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible. Let ๐ฅ cm be the length of a side of the removed square Hence, Length after removing = 18 โ ๐ฅ โ ๐ฅ = 18 โ 2๐ฅ Breadth after removing = 18 โ ๐ฅ โ ๐ฅ = 18 โ 2๐ฅ Height of the box = ๐ฅ We need to maximize volume of box Let V be the volume of a box Volume of a cuboid = l ร b ร h V(๐ฅ)=(18โ2๐ฅ)(18โ2๐ฅ)(๐ฅ) V(๐ฅ)=(18โ2๐ฅ)^2 ๐ฅ Finding vโ(๐ฅ) Vโ (๐ฅ) = ๐((18โ2๐ฅ)^2 ๐ฅ)/๐๐ฅ Vโ (๐ฅ) = ๐((18โ2๐ฅ)^2 )/๐๐ฅ . ๐ฅ + ๐(๐ฅ)/๐๐ฅ . (18โ2๐ฅ)^2 Using product rule . as (๐ข๐ฃ)^โฒ=๐ข^โฒ ๐ฃ+๐ฃ^โฒ ๐ข 18 โ 2๐ฅ = 0 2๐ฅ = 18 ๐ฅ = 18/2= 9 18 โ 6๐ฅ = 0 6๐ฅ = 18 ๐ฅ = 18/6= 3 If ๐ฅ = 9 Breadth of a box = 18 โ 2๐ฅ = 18 โ 2(9) = 18 โ 18 = 0 Since, breadth cannot be zero, โ x = 9 is not possible Hence ๐ฅ = 3 only Now finding Vโโ (๐ฅ) Vโ(๐ฅ)=(18โ2๐ฅ)(18โ6๐ฅ) Vโ (๐ฅ)=๐(18โ2๐ฅ)/๐๐ฅ.(18โ6๐ฅ)+๐(18โ6๐ฅ)/๐๐ฅ.(18โ2๐ฅ) = (0 โ2)(18โ6๐ฅ)+(โ6)(18โ2๐ฅ) = โ2(18โ6๐ฅ)โ6(18โ2๐ฅ) Using product .rule as (๐ข๐ฃ)^โฒ=๐ข^โฒ ๐ฃ+๐ฃ^โฒ ๐ข = โ36 + 12x โ 108 + 12x = 24x โ 144 Putting ๐ฅ = 3 Vโโ(3) = 24 ร 3 โ 144 = โ 72 < 0 Since Vโโ(๐ฅ)<0 for ๐ฅ=3 โด ๐ฅ = 3 is point of Maxima V(x) "is Maximum " at ๐ฅ=3 Hence, 3 cm side of the square to be cut off so that the volume of the box is maximum

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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 9 years. He provides courses for Maths and Science at Teachoo.