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Example 51 - Manufacturer can sell x items at price (5 - x/100)

Example 51 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 51 - Chapter 6 Class 12 Application of Derivatives - Part 3
Example 51 - Chapter 6 Class 12 Application of Derivatives - Part 4
Example 51 - Chapter 6 Class 12 Application of Derivatives - Part 5

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Transcript

Example 51 Manufacturer can sell π‘₯ items at a price of rupees (5βˆ’π‘₯/100) each. The cost price of π‘₯ items is Rs (π‘₯/5+500) Find the number of items he should sell to earn maximum profit.Let S(𝒙) be the Selling Price of π‘₯ items. & C(𝒙) be the cost Price of π‘₯ item. Given Manufacture sell π‘₯ items at a price of rupees (5βˆ’π‘₯/100) each S(𝒙) = π‘₯ Γ— (5βˆ’π‘₯/100) = 5𝒙 – π’™πŸ/𝟏𝟎𝟎 Also given Cost of x items is Rs. (π‘₯/5+500) C(π‘₯) = π‘₯/5 + 500 We need to maximize profit Let P(π‘₯) be the profit Profit = Selling price – cost price P(π‘₯) = S(π‘₯) – C(π‘₯) P(π‘₯) = (5π‘₯βˆ’π‘₯2/100)βˆ’(π‘₯/5+500) = 5π‘₯ – π‘₯2/100βˆ’ π‘₯/5 – 500 = (25π‘₯ βˆ’ π‘₯)/5 – π‘₯2/100 –500 = 24π‘₯/5βˆ’π‘₯2/100 "– 500" Hence, P(π‘₯) = 24π‘₯/5βˆ’π‘₯2/100βˆ’500 Diff w.r.t x P’(π‘₯) = 𝑑(24/5 π‘₯ βˆ’ π‘₯^2/100 βˆ’ 500)/𝑑π‘₯ P’(π‘₯) = 24/5βˆ’2π‘₯/100βˆ’0 P’(π‘₯) = 24/5βˆ’π‘₯/50 Putting P’(π‘₯) = 0 24/5βˆ’π‘₯/50 = 0Putting P’(𝒙) = 0 24/5βˆ’π‘₯/50 = 0 (βˆ’π‘₯)/50=(βˆ’24)/5 π‘₯ = (βˆ’24)/5 Γ— βˆ’50 π‘₯ = 240 Finding P’’(𝒙) P’(π‘₯) = 24/5 – π‘₯/50 Diff w.r.t π‘₯ P’’(π‘₯) = 𝑑(24/5 βˆ’ π‘₯/50)/𝑑π‘₯ = 0 – 1/50 = (βˆ’1)/50 < 0 Since P’’(𝒙) < 0 at π‘₯ = 240 ∴ π‘₯ = 240 is point of maxima Thus, P(π‘₯) 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 has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.