Learn Intergation from Davneet Sir - Live lectures starting soon!

Examples

Example 1

Example 2

Example 3

Example 4 Important

Example 5 You are here

Example 6

Example 7

Example 8 Important

Example 9 Important

Example 10

Example 11 Important

Example 12

Example 13 Important

Example 14 Deleted for CBSE Board 2023 Exams

Example 15 Deleted for CBSE Board 2023 Exams

Example 16 Deleted for CBSE Board 2023 Exams

Example 17 Important Deleted for CBSE Board 2023 Exams

Example 18 Deleted for CBSE Board 2023 Exams

Example 19 Deleted for CBSE Board 2023 Exams

Example 20 Deleted for CBSE Board 2023 Exams

Example 21 Deleted for CBSE Board 2023 Exams

Example 22 Deleted for CBSE Board 2023 Exams

Example 23 Deleted for CBSE Board 2023 Exams

Example 24

Example 25

Example 26

Example 27

Example 28 Important

Example 29

Example 30 Important

Example 31

Example 32 Important

Example 33 Important

Example 34

Example 35 Important

Example 36

Example 37 Important

Example 38 Important

Example 39

Example 40 Important

Example 41 Important

Example 42 Important

Example 43 Important

Example 44 Important

Example 45 Important Deleted for CBSE Board 2023 Exams

Example 46 Important Deleted for CBSE Board 2023 Exams

Example 47 Important

Example 48 Important

Example 49

Example 50 Important

Example 51

Chapter 6 Class 12 Application of Derivatives

Serial order wise

Last updated at April 15, 2021 by Teachoo

Example 5 The total cost C(x) in Rupees, associated with the production of x units of an item is given by C(x) = 0.005 x3 – 0.02 x2 + 30x + 5000 Find the marginal cost when 3 units are produced, where by marginal cost we mean the instantaneous rate of change of total cost at any level of output.Marginal cost is the rate of change of total cost w.r.t output (unit produced) Let MC be marginal cost ∴ MC = 𝒅𝑪/𝒅𝒙 It is given C(𝑥) = 0.005𝑥3 – 0.02𝑥2 + 30𝑥 + 5000 We need to find marginal cost when 3 unit produced i.e. need to find MC = 𝑑𝐶/𝑑𝑥 at 𝑥 = 3 Now, MC = 𝑑𝐶/𝑑𝑥 MC = 𝑑(0.05𝑥^3 − 0.02𝑥2 + 30𝑥 + 5000)/𝑑𝑥 MC = 𝑑(0.05𝑥3)/𝑑𝑥 – (𝑑(0.02𝑥2))/𝑑𝑥 + (𝑑 (30𝑥))/𝑑𝑥 + (𝑑 (5000))/𝑑𝑥 MC = 0.05 × 3x2 – 0.02 × 2x + 30 + 0 MC = 0.015x2 – 0.04x + 30 We need MC at 𝒙 = 3 Putting 𝑥= 3 MC = 0.015(32) – 0.04 (3) + 30 MC = 0.015 × 9 – 0.04 × 3 + 30 MC = 0.135 – 0.12 × 3 + 30 MC = 0.015 + 30 MC = 30.015 Hence, Required marginal cost is Rs. 30.02 (nearly)