# Ex 6.3,17 - Chapter 6 Class 12 Application of Derivatives

Last updated at April 16, 2024 by Teachoo

Ex 6.3

Ex 6.3, 1 (i)
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Ex 6.3, 1 (ii)

Ex 6.3, 1 (iii) Important

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Ex 6.3,17 You are here

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Ex 6.3, 27 (MCQ)

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Ex 6.3,29 (MCQ)

Last updated at April 16, 2024 by Teachoo

Ex 6.3, 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 š„ cm be the length of a side of the removed square Hence, Length after removing = 18 ā š„ ā š„ = 18 ā 2š„ Breadth after removing = 18 ā š„ ā š„ = 18 ā 2š„ Height of the box = š„ We need to maximize volume of box Let V be the volume of a box Volume of a cuboid = l Ć b Ć h V(š„)=(18ā2š„)(18ā2š„)(š„) V(š„)=(18ā2š„)^2 š„ Finding vā(š„) Vā (š„) = š((18ā2š„)^2 š„)/šš„ Vā (š„) = š((18ā2š„)^2 )/šš„ . š„ + š(š„)/šš„ . (18ā2š„)^2 = (2(18ā2š„).š(18ā2š„)/šš„)š„+(1.(18ā2š„)^2 ) = (2(18ā2š„).(0ā2))š„+ (18ā2š„)^2 = ā4(18ā2š„)š„+(18ā2š„)^2 = (18ā2š„) [ā4š„+(18ā2š„)] = (18ā2š„)(18ā6š„) Putting Vā(š„) = 0 So, š„ = 3, 9 If š„ = 9 Breadth of a box = 18 ā 2š„ = 18 ā 2(9) = 18 ā 18 = 0 Since, breadth cannot be zero, ā x = 9 is not possible Hence š„ = 3 only Now finding Vāā (š„) Vā(š„)=(18ā2š„)(18ā6š„) Vā (š„)=š(18ā2š„)/šš„.(18ā6š„)+š(18ā6š„)/šš„.(18ā2š„) = (0 ā2)(18ā6š„)+(ā6)(18ā2š„) = ā2(18ā6š„)ā6(18ā2š„) = ā36 + 12x ā 108 + 12x = 24x ā 144 Putting š„ = 3 Vāā(3) = 24 Ć 3 ā 144 = ā 72 < 0 Since Vāā(š„)<0 for š„=3 ā“ š„ = 3 is point of Maxima V(x) "is Maximum " at š„=3 Hence, 3 cm side of the square to be cut off so that the volume of the box is maximum