# 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

Examples

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6

Example 7

Example 8

Example 9

Example 10

Example 11

Example 12

Example 13

Example 14

Example 15

Example 16

Example 17

Example 18

Example 19

Example 20

Example 21

Example 22

Example 23

Example 24

Example 25

Example 26

Example 27

Example 28

Example 29

Example 30

Example 31

Example 32

Example 33

Example 34

Example 35 Important

Example 36

Example 37 Important

Example 38 Important

Example 39

Example 40 Important

Example 41

Example 42

Example 43

Example 44

Example 45

Example 46 Important

Example 47 Important

Example 48

Example 49

Example 50 You are here

Example 51

Chapter 6 Class 12 Application of Derivatives

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.