# Ex 6.5,17 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 15, 2021 by Teachoo

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Ex 6.5,17 You are here

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Last updated at April 15, 2021 by Teachoo

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 π₯ cm be the length of a side of the removed square Hence, Length after removing = 18 β π₯ β π₯ = 18 β 2π₯ Breadth after removing = 18 β π₯ β π₯ = 18 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 V(π₯)=(18β2π₯)(18β2π₯)(π₯) V(π₯)=(18β2π₯)^2 π₯ Finding vβ(π₯) Vβ (π₯) = π((18β2π₯)^2 π₯)/ππ₯ Vβ (π₯) = π((18β2π₯)^2 )/ππ₯ . π₯ + π(π₯)/ππ₯ . (18β2π₯)^2 Using product rule . as (π’π£)^β²=π’^β² π£+π£^β² π’ = (2(18β2π₯).π(18β2π₯)/ππ₯)π₯+(1.(18β2π₯)^2 ) = (2(18β2π₯).(0β2))π₯+ (18β2π₯)^2 = β4(18β2π₯)π₯+(18β2π₯)^2 = (18β2π₯) [β4π₯+(18β2π₯)] = (18β2π₯)(18β6π₯) Putting Vβ(π₯) = 0 So, π₯ = 3, 9 18 β 2π₯ = 0 2π₯ = 18 π₯ = 18/2= 9 18 β 6π₯ = 0 6π₯ = 18 π₯ = 18/6= 3 If π₯ = 9 Breadth of a box = 18 β 2π₯ = 18 β 2(9) = 18 β 18 = 0 Since, breadth cannot be zero, β x = 9 is not possible Hence π₯ = 3 only Now finding Vββ (π₯) Vβ(π₯)=(18β2π₯)(18β6π₯) Vβ (π₯)=π(18β2π₯)/ππ₯.(18β6π₯)+π(18β6π₯)/ππ₯.(18β2π₯) = (0 β2)(18β6π₯)+(β6)(18β2π₯) Using product .rule as (π’π£)^β²=π’^β² π£+π£^β² π’ = β2(18β6π₯)β6(18β2π₯) = β36 + 12x β 108 + 12x = 24x β 144 Putting π₯ = 3 Vββ(3) = 24 Γ 3 β 144 = β 72 < 0 Since Vββ(π₯)<0 for π₯=3 β΄ π₯ = 3 is point of Maxima V(x) "is Maximum " at π₯=3 Hence, 3 cm side of the square to be cut off so that the volume of the box is maximum v