# Example 50 - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

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 Length after removing = 8 – 𝑥 – 𝑥 = 8 – 2𝑥 Breadth after removing = 3 – 𝑥 – 𝑥 = 3 – 2𝑥 Height of the box = 𝑥 We need to maximize volume of box Let V be the volume of a box V = (length) (breadth) (height) = 8−2𝑥3−2𝑥𝑥 = 8−2𝑥3𝑥−2𝑥2 = 83𝑥−2𝑥2 – 2x 3𝑥−2𝑥2 = 24𝑥 – 16x2 – 6𝑥2 + 4𝑥3 = 4𝑥3 – 22𝑥2 + 24𝑥 Now, 𝑉𝑥 = 4𝑥3 – 22𝑥2 + 24𝑥 Diff w.r.t. x 𝑉′𝑥 = 𝑑4𝑥3 – 22𝑥2 + 24𝑥 𝑑𝑥 𝑉′𝑥 = 4 × 3x2 – 22 × 2𝑥 + 24 𝑉′𝑥 = 12𝑥2 – 44𝑥 + 24 𝑉′𝑥 = 43𝑥2−11𝑥+6 Putting 𝑉′𝑥 = 0 43𝑥2−11𝑥+6 = 0 3𝑥2−11𝑥+6 = 0 3𝑥2 –9𝑥 – 2𝑥 + 6 = 0 3𝑥𝑥−3 –2 𝑥−3 = 0 3𝑥−2𝑥−3= 0 So, 𝑥=23 & 𝑥=3 If 𝑥 = 3 breadth of a box = 3 – 2𝑥 = 3 – 2(3) = 3 – 6 = –3 Since, breadth cannot be negative, ⇒ x = 3 is not possible Hence 𝑥 = 23 only Finding 𝑉’’𝑥 𝑉’𝑥 = 43𝑥2−11𝑥+6 Diff w.r.t 𝑥 𝑉’’𝑥 = 𝑑4(3𝑥2 − 11𝑥 + 6𝑑𝑥 𝑉’’𝑥 = 4 3×2𝑥−11 𝑉’’𝑥 = 4 6𝑥−11 Putting x = 23 in 𝑉’’𝑥 𝑉’’23=4623−11 = 4 4−11= –28 < 0 ∴ 𝑉’’𝑥 < 0 at 𝑥 = 23 ⇒ 𝑥 = 23 is point of maxima ⇒ 𝑉𝑥 is largest when 𝑥 = 23 Largest volume is 𝑉𝑥 = x3−2𝑥 8−2𝑥 at x = 23 𝑉23 = 233−223 8−223 = 23 3−438−43 = 23 9 − 4324 − 43 = 23 53203 = 20027 Since dimension of volume is m3 Largest volume is 𝟐𝟎𝟎𝟐𝟕 m3

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Example 50 You are here

Example 51

Chapter 6 Class 12 Application of Derivatives

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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 9 years. He provides courses for Maths and Science at Teachoo.