Last updated at April 19, 2021 by Teachoo

Transcript

Example 51 Manufacturer can sell ๐ฅ items at a price of rupees (5โ๐ฅ/100) each. The cost price of ๐ฅ items is Rs (๐ฅ/5+500) Find the number of items he should sell to earn maximum profit.Let S(๐) be the Selling Price of ๐ฅ items. & C(๐) be the cost Price of ๐ฅ item. Given Manufacture sell ๐ฅ items at a price of rupees (5โ๐ฅ/100) each S(๐) = ๐ฅ ร (5โ๐ฅ/100) = 5๐ โ ๐๐/๐๐๐ Also given Cost of x items is Rs. (๐ฅ/5+500) C(๐ฅ) = ๐ฅ/5 + 500 We need to maximize profit Let P(๐ฅ) be the profit Profit = Selling price โ cost price P(๐ฅ) = S(๐ฅ) โ C(๐ฅ) P(๐ฅ) = (5๐ฅโ๐ฅ2/100)โ(๐ฅ/5+500) = 5๐ฅ โ ๐ฅ2/100โ ๐ฅ/5 โ 500 = (25๐ฅ โ ๐ฅ)/5 โ ๐ฅ2/100 โ500 = 24๐ฅ/5โ๐ฅ2/100 "โ 500" Hence, P(๐ฅ) = 24๐ฅ/5โ๐ฅ2/100โ500 Diff w.r.t x Pโ(๐ฅ) = ๐(24/5 ๐ฅ โ ๐ฅ^2/100 โ 500)/๐๐ฅ Pโ(๐ฅ) = 24/5โ2๐ฅ/100โ0 Pโ(๐ฅ) = 24/5โ๐ฅ/50 Putting Pโ(๐ฅ) = 0 24/5โ๐ฅ/50 = 0Putting Pโ(๐) = 0 24/5โ๐ฅ/50 = 0 (โ๐ฅ)/50=(โ24)/5 ๐ฅ = (โ24)/5 ร โ50 ๐ฅ = 240 Finding Pโโ(๐) Pโ(๐ฅ) = 24/5 โ ๐ฅ/50 Diff w.r.t ๐ฅ Pโโ(๐ฅ) = ๐(24/5 โ ๐ฅ/50)/๐๐ฅ = 0 โ 1/50 = (โ1)/50 < 0 Since Pโโ(๐) < 0 at ๐ฅ = 240 โด ๐ฅ = 240 is point of maxima Thus, P(๐ฅ) is maximum when ๐ = 240 Hence the manufacturer can earn maximum profit if he sells 240 items.

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Example 51 You are here

Chapter 6 Class 12 Application of Derivatives (Term 1)

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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 10 years. He provides courses for Maths and Science at Teachoo.