Ex 6.3,18 - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Ex 6.3
Ex 6.3, 1 (ii)
Ex 6.3, 1 (iii) Important
Ex 6.3, 1 (iv)
Ex 6.3, 2 (i)
Ex 6.3, 2 (ii) Important
Ex 6.3, 2 (iii)
Ex 6.3, 2 (iv) Important
Ex 6.3, 2 (v) Important
Ex 6.3, 3 (i)
Ex 6.3, 3 (ii)
Ex 6.3, 3 (iii)
Ex 6.3, 3 (iv) Important
Ex 6.3, 3 (v)
Ex 6.3, 3 (vi)
Ex 6.3, 3 (vii) Important
Ex 6.3, 3 (viii)
Ex 6.3, 4 (i)
Ex 6.3, 4 (ii) Important
Ex 6.3, 4 (iii)
Ex 6.3, 5 (i)
Ex 6.3, 5 (ii)
Ex 6.3, 5 (iii) Important
Ex 6.3, 5 (iv)
Ex 6.3,6
Ex 6.3,7 Important
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Ex 6.3,9 Important
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Ex 6.3,11 Important
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Ex 6.3,18 Important You are here
Ex 6.3,19 Important
Ex 6.3, 20 Important
Ex 6.3,21
Ex 6.3,22 Important
Ex 6.3,23 Important
Ex 6.3,24 Important
Ex 6.3,25 Important
Ex 6.3, 26 Important
Ex 6.3, 27 (MCQ)
Ex 6.3,28 (MCQ) Important
Ex 6.3,29 (MCQ)
Last updated at April 16, 2024 by Teachoo
Ex 6.3, 18 A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum ?Let š„ be the length of a side of the removed square Thus, Length after removing = 45 ā š„ āš„ = 45 ā 2š„ Breadth after removing = 24 āš„ āš„ = 24 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 = (45ā2š„)(24ā2š„)(š„) = [45(24ā2š„)ā2š„(24ā2š„)]š„ = (1080ā90š„ā48š„+4š„^2 )š„ = 1080š„ā90š„^2ā48š„^2+4š„^3 = 1080š„ā138š„^2+4š„^3 = 2(540š„ā69š„^2+2š„^3 ) = 2(2š„^3ā69š„^2+540š„) V = 2(2š„^3ā69š„^2+540) Diff w.r.t š„ Vā(š„)=2š[2š„^3ā 69š„^2+ 540š„]/šš„ = 2[6š„^2ā69 Ć2š„+540] = 2[6š„^2ā138š„+540] = 2 Ć 6[š„^2ā23š„+90] = 12[š„^2ā23š„+90] Putting Vā(š„)=0 12(š„^2ā23š„+90)=0 š„^2ā23š„+90=0 š„^2ā5š„ā18š„+90=0 š„(š„ā5)ā18(š„ā5)=0 (š„ā18)(š„ā5)=0 So, x = 18 & x = 5 If š„ = 18 Breadth of a box = 24 ā 2š„ = 24 ā 2(18) = 24 ā 36 = ā12 Since, breadth cannot be negative, ā x = 18 is not possible Hence š„ = 5 only Finding Vāā(š„) Vā(š„)=12(š„^2ā23š„+90) Vāā(š„)=12(2š„ā23) Putting š„=5 Vāā(5)=12(2(5)ā23)= 12(10ā23)= 12(ā 13)= ā 156 Vāā(š„)<0 when š„=5 Thus, V(š„) is maximum at š„=5 ā“ Square of side 5 cm is cut off from each Corner