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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.