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Example 51 - Manufacturer can sell x items at price (5 - x/100) - Minima/ maxima (statement questions) - Number questions

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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Example 51 Manufacturer can sell 𝑥 items at a price of rupees ﷐5−﷐𝑥﷮100﷯﷯ each. The cost price of 𝑥 items is Rs ﷐﷐𝑥﷮5﷯+500﷯ Find the number of items he should sell to earn maximum profit. Let S﷐𝑥﷯ be the selling price of 𝑥 items. & C﷐𝑥﷯ be the cost price of 𝑥 item. Given Manufacture sell 𝑥 items at a price of rupees ﷐5−﷐𝑥﷮100﷯﷯ each i.e. S﷐𝑥﷯ = 𝑥 × ﷐5−﷐𝑥﷮100﷯﷯ = 5𝑥 – ﷐𝑥2﷮100﷯ Also given Cost of x items is Rs. ﷐﷐𝑥﷮5﷯+500﷯ i.e. C﷐𝑥﷯ = ﷐𝑥﷮5﷯ + 500 We need to maximize profit let P﷐𝑥﷯ be the profit P﷐𝑥﷯ = selling price – cost price P﷐𝑥﷯ = S﷐𝑥﷯ – C﷐𝑥﷯ P﷐𝑥﷯ = ﷐5𝑥−﷐𝑥2﷮100﷯﷯−﷐﷐𝑥﷮5﷯+500﷯ = 5𝑥 – ﷐𝑥2﷮100﷯− ﷐𝑥﷮5﷯ – 500 = ﷐25𝑥−𝑥﷮5﷯ – ﷐𝑥2﷮100﷯ –500 = ﷐24𝑥﷮5﷯−﷐𝑥2﷮100﷯−500 Hence, P﷐𝑥﷯ = ﷐24𝑥﷮5﷯−﷐𝑥2﷮100﷯−500 Diff w.r.t x P’﷐𝑥﷯ = ﷐𝑑﷐﷐24﷮5﷯ 𝑥 − ﷐﷐𝑥﷮2﷯﷮100﷯ − 500﷯﷮𝑑𝑥﷯ P’﷐𝑥﷯ = ﷐24﷮5﷯−﷐2𝑥﷮100﷯−0 P’﷐𝑥﷯ = ﷐24﷮5﷯−﷐𝑥﷮50﷯ Putting P’﷐𝑥﷯ = 0 ﷐24﷮5﷯−﷐𝑥﷮50﷯ = 0 ﷐−𝑥﷮50﷯=﷐−24﷮5﷯ 𝑥 = ﷐−24﷮5﷯ ×−50 𝑥 = 240 Finding P’’﷐𝑥﷯ P’﷐𝑥﷯ = ﷐24﷮5﷯ – ﷐𝑥﷮50﷯ Diff w.r.t 𝑥 P’’﷐𝑥﷯ = ﷐𝑑﷐﷐24﷮5﷯ − ﷐𝑥﷮50﷯﷯﷮𝑑𝑥﷯ = 0 – ﷐1﷮50﷯ = ﷐−1﷮50﷯ Putting 𝑥 = 240 ﷐𝑃﷮′′﷯﷐240﷯ = ﷐−1﷮50﷯ < 0 ⇒ P’’﷐𝑥﷯ < 0 at 𝑥 = 240 ⇒ 𝑥 = 240 is point of maxima Thus, P﷐𝑥﷯ is maximum when 𝑥 = 240 Hence the manufacturer can earn maximum profit if he sells 240 items.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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