Ex 6.1,15 - Chapter 6 Class 12 Application of Derivatives
Last updated at August 13, 2026 by Teachoo
Finding rate of change
Example 1
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Chapter 6 Class 12 Application of Derivatives
Concept wise
-
Finding rate of change
- To show increasing/decreasing in whole domain
- To show increasing/decreasing in intervals
- Find intervals of increasing/decreasing
- Finding slope of tangent/normal
- Finding point when tangent is parallel/ perpendicular
- Finding equation of tangent/normal when point and curve is given
- Finding equation of tangent/normal when slope and curve are given
- Finding approximate value of numbers
- Finding approximate value of function
- Finding approximate value- Statement questions
- Finding minimum and maximum values from graph
- Local maxima and minima
- Minima/ maxima (statement questions) - Number questions
- Minima/ maxima (statement questions) - Geometry questions
- Absolute minima/maxima
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Transcript
Ex 6.1, 15 The total cost C(š„) in Rupees associated with the production of š„ units of an item is given by š¶(š„) = 0.007š„^3 ā 0.003š„2 + 15š„ + 4000. Find the marginal cost when 17 units are produced.Since Marginal Cost is Rate of change in Total Cost w.r.t No of units produced Let MC be marginal cost So, MC = š šŖ/š š Given, Total Cost = C(š„) =0.007š„^3ā0.003š„^2+15š„+4000" " We need to find marginal Cost when 17 units are produced i.e. MC when š„ = 17 MC = š(š¶(š„))/šš„ MC = š(0.007š„^3 ā 0.003š„^2 + 15š„ + 4000)/šš„ MC = š(0.007š„^3 )/šš„ ā š(0.003š„^2 )/šš„+ š(15š„)/šš„+ š(4000)/šš„ MC = 0.007 š(š„^3 )/šš„ ā0.003 š(š„^2 )/šš„ + 15š(š„)/šš„+0 MC = 0.007 Ć 3š„^2ā0.003 Ć2š„+15 MC = 0.021š„^2ā0.006š„+15 We need to find MC when š„=17 Putting š=šš MC = 0.021(17)^2ā0.006(17)+15 MC = 6.069 ā 0.102 + 15 MC = 20.967 Hence, the Required Marginal cost is Rs. 20.967
