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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise


Example 51 Manufacturer can sell ๐‘ฅ items at a price of rupees (5โˆ’๐‘ฅ/100) each. The cost price of ๐‘ฅ items is Rs (๐‘ฅ/5+500) Find the number of items he should sell to earn maximum profit.Let S(๐’™) be the Selling Price of ๐‘ฅ items. & C(๐’™) be the cost Price of ๐‘ฅ item. Given Manufacture sell ๐‘ฅ items at a price of rupees (5โˆ’๐‘ฅ/100) each S(๐’™) = ๐‘ฅ ร— (5โˆ’๐‘ฅ/100) = 5๐’™ โ€“ ๐’™๐Ÿ/๐Ÿ๐ŸŽ๐ŸŽ Also given Cost of x items is Rs. (๐‘ฅ/5+500) C(๐‘ฅ) = ๐‘ฅ/5 + 500 We need to maximize profit Let P(๐‘ฅ) be the profit Profit = Selling price โ€“ cost price P(๐‘ฅ) = S(๐‘ฅ) โ€“ C(๐‘ฅ) P(๐‘ฅ) = (5๐‘ฅโˆ’๐‘ฅ2/100)โˆ’(๐‘ฅ/5+500) = 5๐‘ฅ โ€“ ๐‘ฅ2/100โˆ’ ๐‘ฅ/5 โ€“ 500 = (25๐‘ฅ โˆ’ ๐‘ฅ)/5 โ€“ ๐‘ฅ2/100 โ€“500 = 24๐‘ฅ/5โˆ’๐‘ฅ2/100 "โ€“ 500" Hence, P(๐‘ฅ) = 24๐‘ฅ/5โˆ’๐‘ฅ2/100โˆ’500 Diff w.r.t x Pโ€™(๐‘ฅ) = ๐‘‘(24/5 ๐‘ฅ โˆ’ ๐‘ฅ^2/100 โˆ’ 500)/๐‘‘๐‘ฅ Pโ€™(๐‘ฅ) = 24/5โˆ’2๐‘ฅ/100โˆ’0 Pโ€™(๐‘ฅ) = 24/5โˆ’๐‘ฅ/50 Putting Pโ€™(๐‘ฅ) = 0 24/5โˆ’๐‘ฅ/50 = 0Putting Pโ€™(๐’™) = 0 24/5โˆ’๐‘ฅ/50 = 0 (โˆ’๐‘ฅ)/50=(โˆ’24)/5 ๐‘ฅ = (โˆ’24)/5 ร— โˆ’50 ๐‘ฅ = 240 Finding Pโ€™โ€™(๐’™) Pโ€™(๐‘ฅ) = 24/5 โ€“ ๐‘ฅ/50 Diff w.r.t ๐‘ฅ Pโ€™โ€™(๐‘ฅ) = ๐‘‘(24/5 โˆ’ ๐‘ฅ/50)/๐‘‘๐‘ฅ = 0 โ€“ 1/50 = (โˆ’1)/50 < 0 Since Pโ€™โ€™(๐’™) < 0 at ๐‘ฅ = 240 โˆด ๐‘ฅ = 240 is point of maxima Thus, P(๐‘ฅ) is maximum when ๐’™ = 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 has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.