# Ex 6.5,17 - Chapter 6 Class 12 Application of Derivatives

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

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 š„ cm be the length of a side of the removed square 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 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 side cut off be š„ cm Length after cutting = 18 ā š„ ā š„ = 18 ā 2š„ Breadth after cutting = 18 ā š„ ā š„ = 18 ā 2š„ Height = š„ Volume = Length Ć Breadth Ć Height V =ļ·18ā2š„ļ·Æ ļ·18ā2š„ļ·Æ š„ Let V ļ·š„ļ·Æ = ļ·18ā2š„ļ·Æ ļ·18ā2š„ļ·Æ š„ We need to find Max volume as possible Hence we need to find Maximum value of V Volume of a cuboid = l Ć b Ć h Vļ·š„ļ·Æ=ļ·18ā2š„ļ·Æļ·18ā2š„ļ·Æļ·š„ļ·Æ Vļ·š„ļ·Æ=ļ·ļ·18ā2š„ļ·Æļ·®2ļ·Æš„ Finding vāļ·š„ļ·Æ Vā ļ·š„ļ·Æ = ļ·šļ·ļ·ļ·18ā2š„ļ·Æļ·®2ļ·Æš„ļ·Æļ·®šš„ļ·Æ Vā ļ·š„ļ·Æ = ļ·šļ·ļ·ļ·18ā2š„ļ·Æļ·®2ļ·Æļ·Æļ·®šš„ļ·Æ . š„ + ļ·šļ·š„ļ·Æļ·®šš„ļ·Æ . ļ·ļ·18ā2š„ļ·Æļ·®2ļ·Æ = ļ·2ļ·18ā2š„ļ·Æ.ļ·šļ·18ā2š„ļ·Æļ·®šš„ļ·Æļ·Æš„+ļ·1.ļ·ļ·18ā2š„ļ·Æļ·®2ļ·Æļ·Æ = ļ·2ļ·18ā2š„ļ·Æ.ļ·0ā2ļ·Æļ·Æš„+ ļ·ļ·18ā2š„ļ·Æļ·®2ļ·Æ = ā4ļ·18ā2š„ļ·Æš„+ļ·ļ·18ā2š„ļ·Æļ·®2ļ·Æ = ļ·18ā2š„ļ·Æ ļ·ā4š„+ļ·18ā2š„ļ·Æļ·Æ = ļ·18ā2š„ļ·Æļ·18ā6š„ļ·Æ Putting Vāļ·š„ļ·Æ = 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 ļ·š„ļ·Æ Vāļ·š„ļ·Æ=ļ·18ā2š„ļ·Æļ·18ā6š„ļ·Æ Vā ļ·š„ļ·Æ=ļ·šļ·18ā2š„ļ·Æļ·®šš„ļ·Æ.ļ·18ā6š„ļ·Æ+ļ·šļ·18ā6š„ļ·Æļ·®šš„ļ·Æ.ļ·18ā2š„ļ·Æ = ļ·0 ā2ļ·Æļ·18ā6š„ļ·Æ+ļ·ā6ļ·Æļ·18ā2š„ļ·Æ = ā2ļ·18ā6š„ļ·Æā6ļ·18ā2š„ļ·Æ = ā36 + 12x ā 108 + 12x = 24x ā 144 Putting š„ = 3 Vāā(3) = 23 Ć 3 ā 144 = ā 72 < 0 Since Vāāļ·š„ļ·Æ<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

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Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.