# Ex 6.5,18 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 15, 2021 by

Last updated at April 15, 2021 by

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Ex 6.5, 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 π₯ be the length of a side of the removed square Thus, Length after removing = 45 β π₯ βπ₯ = 45 β 2π₯ Breadth after removing = 24 βπ₯ βπ₯ = 24 β 2π₯ Height of the box = π₯ We need to maximize volume of box Let V be the volume of a box Volume of a cuboid = l Γ b Γ h = (45β2π₯)(24β2π₯)(π₯) = [45(24β2π₯)β2π₯(24β2π₯)]π₯ = (1080β90π₯β48π₯+4π₯^2 )π₯ = 1080π₯β90π₯^2β48π₯^2+4π₯^3 = 1080π₯β138π₯^2+4π₯^3 = 2(540π₯β69π₯^2+2π₯^3 ) = 2(2π₯^3β69π₯^2+540π₯) V = 2(2π₯^3β69π₯^2+540) Diff w.r.t π₯ Vβ(π₯)=2π[2π₯^3β 69π₯^2+ 540π₯]/ππ₯ = 2[6π₯^2β69 Γ2π₯+540] = 2[6π₯^2β138π₯+540] = 2 Γ 6[π₯^2β23π₯+90] = 12[π₯^2β23π₯+90] Putting Vβ(π₯)=0 12(π₯^2β23π₯+90)=0 π₯^2β23π₯+90=0 π₯^2β5π₯β18π₯+90=0 π₯(π₯β5)β18(π₯β5)=0 (π₯β18)(π₯β5)=0 So, x = 18 & x = 5 If π₯ = 18 Breadth of a box = 24 β 2π₯ = 24 β 2(18) = 24 β 36 = β12 Since, breadth cannot be negative, β x = 18 is not possible Hence π₯ = 5 only Finding Vββ(π₯) Vβ(π₯)=12(π₯^2β23π₯+90) Vββ(π₯)=12(2π₯β23) Putting π₯=5 Vββ(5)=12(2(5)β23)= 12(10β23)= 12(β 13)= β 156 Vββ(π₯)<0 when π₯=5 Thus, V(π₯) is maximum at π₯=5 β΄ Square of side 5 cm is cut off from each Corner

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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.