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Example 50 - An open topped box is to be constructed by - Examples

Example 50 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 50 - Chapter 6 Class 12 Application of Derivatives - Part 3
Example 50 - Chapter 6 Class 12 Application of Derivatives - Part 4
Example 50 - Chapter 6 Class 12 Application of Derivatives - Part 5
Example 50 - Chapter 6 Class 12 Application of Derivatives - Part 6

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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 𝒙 m be the length of a side of the removed square Hence, Length after removing = 8 – π‘₯ – π‘₯ = 8 – 2𝒙 Breadth after removing = 3 – π‘₯ – π‘₯ = 3 – 2𝒙 Height of the box = 𝒙 We need to maximize volume of box Let V be the volume of a box V = Length Γ— Breadth Γ— Height) = (8βˆ’2π‘₯)(3βˆ’2π‘₯)(π‘₯) = (8βˆ’2π‘₯)(3π‘₯βˆ’2π‘₯2) = 8(3π‘₯βˆ’2π‘₯2) – 2x (3π‘₯βˆ’2π‘₯2) = 24π‘₯ – 16x2 – 6π‘₯2 + 4π‘₯3 = 4𝒙3 – 22𝒙2 + 24𝒙 Now, 𝑉(π‘₯) = 4π‘₯3 – 22π‘₯2 + 24π‘₯ Diff w.r.t. x 𝑉′(π‘₯) = 𝑑(4π‘₯^3 βˆ’ 22π‘₯^2 + 24π‘₯)/𝑑π‘₯ 𝑉′(π‘₯) = 4 Γ— 3x2 – 22 Γ— 2π‘₯ + 24 𝑉′(π‘₯) = 12π‘₯2 – 44π‘₯ + 24 𝑉′(π‘₯) = 4(3π‘₯2βˆ’11π‘₯+6) Putting 𝑽′(𝒙) = 0 4(3π‘₯2βˆ’11π‘₯+6) = 0 3π‘₯2βˆ’11π‘₯+6 = 0 3π‘₯2 –9π‘₯ – 2π‘₯ + 6 = 0 3π‘₯(π‘₯βˆ’3) –2 (π‘₯βˆ’3) = 0 (3π‘₯βˆ’2)(π‘₯βˆ’3)= 0 So, 𝒙=𝟐/πŸ‘ & 𝒙=πŸ‘ If 𝒙 = 3 Breadth of a box = 3 – 2π‘₯ = 3 – 2(3) = 3 – 6 = –3 Since, breadth cannot be negative, ∴ x = 3 is not possible Hence, 𝒙 = 𝟐/πŸ‘ only Finding 𝑽’’(𝒙) 𝑉’(π‘₯) = 4(3π‘₯2βˆ’11π‘₯+6) Diff w.r.t π‘₯ 𝑉’’(π‘₯) = 𝑑(4(3π‘₯^2 βˆ’ 11π‘₯ + 6)/𝑑π‘₯ 𝑉’’(π‘₯) = 4 (3Γ—2π‘₯βˆ’11) 𝑉’’(π‘₯) = 4 (6π‘₯βˆ’11) Putting x = 𝟐/πŸ‘ 𝑽’’(𝟐/πŸ‘)=4(6(2/3)βˆ’11) = 4 (4βˆ’11)= –28 < 0 Since 𝑉’’(π‘₯) < 0 at π‘₯ = 2/3 ∴ π‘₯ = 2/3 is point of maxima Hence, 𝑽(𝒙) is largest when 𝒙 = 𝟐/πŸ‘ Largest volume is 𝑉(π‘₯) = x(3βˆ’2π‘₯) (8βˆ’2π‘₯) 𝑽(𝟐/πŸ‘) = 2/3 (3βˆ’2(2/3)) (8βˆ’2(2/3)) = 2/3 (3βˆ’4/3)(8βˆ’4/3) = 2/3 ((9 βˆ’ 4)/3)((24 βˆ’ 4)/3) = 2/3 (5/3)(20/3) = 200/27 Since dimension of volume is m3 Largest volume is 𝟐𝟎𝟎/πŸπŸ• m3

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.