Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

Last updated at Jan. 7, 2020 by Teachoo

Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

Transcript

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 produce) 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 = (𝑑(𝑐))/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)

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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.