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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


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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.