Last updated at April 19, 2021 by

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Example 50 An open topped box is to be constructed by removing equal squares from each corner of a 3 meter by 8 meter rectangular sheet of aluminum and folding up the sides. Find the volume of the largest such box.Let π m be the length of a side of the removed square Hence, Length after removing = 8 β π₯ β π₯ = 8 β 2π Breadth after removing = 3 β π₯ β π₯ = 3 β 2π Height of the box = π We need to maximize volume of box Let V be the volume of a box V = Length Γ Breadth Γ Height) = (8β2π₯)(3β2π₯)(π₯) = (8β2π₯)(3π₯β2π₯2) = 8(3π₯β2π₯2) β 2x (3π₯β2π₯2) = 24π₯ β 16x2 β 6π₯2 + 4π₯3 = 4π3 β 22π2 + 24π Now, π(π₯) = 4π₯3 β 22π₯2 + 24π₯ Diff w.r.t. x πβ²(π₯) = π(4π₯^3 β 22π₯^2 + 24π₯)/ππ₯ πβ²(π₯) = 4 Γ 3x2 β 22 Γ 2π₯ + 24 πβ²(π₯) = 12π₯2 β 44π₯ + 24 πβ²(π₯) = 4(3π₯2β11π₯+6) Putting π½β²(π) = 0 4(3π₯2β11π₯+6) = 0 3π₯2β11π₯+6 = 0 3π₯2 β9π₯ β 2π₯ + 6 = 0 3π₯(π₯β3) β2 (π₯β3) = 0 (3π₯β2)(π₯β3)= 0 So, π=π/π & π=π If π = 3 Breadth of a box = 3 β 2π₯ = 3 β 2(3) = 3 β 6 = β3 Since, breadth cannot be negative, β΄ x = 3 is not possible Hence, π = π/π only Finding π½ββ(π) πβ(π₯) = 4(3π₯2β11π₯+6) Diff w.r.t π₯ πββ(π₯) = π(4(3π₯^2 β 11π₯ + 6)/ππ₯ πββ(π₯) = 4 (3Γ2π₯β11) πββ(π₯) = 4 (6π₯β11) Putting x = π/π π½ββ(π/π)=4(6(2/3)β11) = 4 (4β11)= β28 < 0 Since πββ(π₯) < 0 at π₯ = 2/3 β΄ π₯ = 2/3 is point of maxima Hence, π½(π) is largest when π = π/π Largest volume is π(π₯) = x(3β2π₯) (8β2π₯) π½(π/π) = 2/3 (3β2(2/3)) (8β2(2/3)) = 2/3 (3β4/3)(8β4/3) = 2/3 ((9 β 4)/3)((24 β 4)/3) = 2/3 (5/3)(20/3) = 200/27 Since dimension of volume is m3 Largest volume is πππ/ππ m3

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About the Author

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.