Examples

Example 1

Example 2

Example 3

Example 4 Important

Example 5

Example 6

Example 7

Example 8 Important

Example 9 Important

Example 10

Example 11 Important

Example 12

Example 13 Important

Example 14

Example 15

Example 16 Important

Example 17

Example 18 Important

Example 19

Example 20 Important

Example 21 Important

Example 22

Example 23 Important

Example 24

Example 25 Important

Example 26 Important

Example 27

Example 28 Important

Example 29 Important

Example 30 Important

Example 31 Important

Example 32 Important

Example 33 Important

Example 34 Important

Example 35

Example 36 Important You are here

Example 37

Question 1 Deleted for CBSE Board 2025 Exams

Question 2 Deleted for CBSE Board 2025 Exams

Question 3 Deleted for CBSE Board 2025 Exams

Question 4 Important Deleted for CBSE Board 2025 Exams

Question 5 Deleted for CBSE Board 2025 Exams

Question 6 Deleted for CBSE Board 2025 Exams

Question 7 Deleted for CBSE Board 2025 Exams

Question 8 Deleted for CBSE Board 2025 Exams

Question 9 Deleted for CBSE Board 2025 Exams

Question 10 Deleted for CBSE Board 2025 Exams

Question 11 Deleted for CBSE Board 2025 Exams

Question 12 Deleted for CBSE Board 2025 Exams

Question 13 Important Deleted for CBSE Board 2025 Exams

Question 14 Important Deleted for CBSE Board 2025 Exams

Last updated at April 16, 2024 by Teachoo

Example 36 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 – 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) = (8−2𝑥)(3−2𝑥)(𝑥) = (8−2𝑥)(3𝑥−2𝑥2) = 8(3𝑥−2𝑥2) – 2x (3𝑥−2𝑥2) = 24𝑥 – 16x2 – 6𝑥2 + 4𝑥3 = 4𝒙3 – 22𝒙2 + 24𝒙 Now, 𝑉(𝑥) = 4𝑥3 – 22𝑥2 + 24𝑥 Diff w.r.t. x 𝑉′(𝑥) = 𝑑(4𝑥^3 − 22𝑥^2 + 24𝑥)/𝑑𝑥 𝑉′(𝑥) = 4 × 3x2 – 22 × 2𝑥 + 24 𝑉′(𝑥) = 12𝑥2 – 44𝑥 + 24 𝑉′(𝑥) = 4(3𝑥2−11𝑥+6) Putting 𝑽′(𝒙) = 0 4(3𝑥2−11𝑥+6) = 0 3𝑥2−11𝑥+6 = 0 3𝑥2 –9𝑥 – 2𝑥 + 6 = 0 3𝑥(𝑥−3) –2 (𝑥−3) = 0 (3𝑥−2)(𝑥−3)= 0 So, 𝒙=𝟐/𝟑 & 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 = 𝟐/𝟑 only Finding 𝑽’’(𝒙) 𝑉’(𝑥) = 4(3𝑥2−11𝑥+6) Diff w.r.t 𝑥 𝑉’’(𝑥) = 𝑑(4(3𝑥^2 − 11𝑥 + 6)/𝑑𝑥 𝑉’’(𝑥) = 4 (3×2𝑥−11) 𝑉’’(𝑥) = 4 (6𝑥−11) Putting x = 𝟐/𝟑 𝑽’’(𝟐/𝟑)=4(6(2/3)−11) = 4 (4−11)= –28 < 0 Since 𝑉’’(𝑥) < 0 at 𝑥 = 2/3 ∴ 𝑥 = 2/3 is point of maxima Hence, 𝑽(𝒙) is largest when 𝒙 = 𝟐/𝟑 Largest volume is 𝑉(𝑥) = x(3−2𝑥) (8−2𝑥) 𝑽(𝟐/𝟑) = 2/3 (3−2(2/3)) (8−2(2/3)) = 2/3 (3−4/3)(8−4/3) = 2/3 ((9 − 4)/3)((24 − 4)/3) = 2/3 (5/3)(20/3) = 200/27 Since dimension of volume is m3 Largest volume is 𝟐𝟎𝟎/𝟐𝟕 m3