Minima/ maxima (statement questions) - Number questions

Chapter 6 Class 12 Application of Derivatives (Term 1)
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Example 51 Manufacturer can sell π₯ items at a price of rupees (5β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(π) be the Selling Price of π₯ items. & C(π) be the cost Price of π₯ item. Given Manufacture sell π₯ items at a price of rupees (5βπ₯/100) each S(π) = π₯ Γ (5βπ₯/100) = 5π β ππ/πππ 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 β cost price P(π₯) = S(π₯) β C(π₯) P(π₯) = (5π₯βπ₯2/100)β(π₯/5+500) = 5π₯ β π₯2/100β π₯/5 β 500 = (25π₯ β π₯)/5 β π₯2/100 β500 = 24π₯/5βπ₯2/100 "β 500" Hence, P(π₯) = 24π₯/5βπ₯2/100β500 Diff w.r.t x Pβ(π₯) = π(24/5 π₯ β π₯^2/100 β 500)/ππ₯ Pβ(π₯) = 24/5β2π₯/100β0 Pβ(π₯) = 24/5βπ₯/50 Putting Pβ(π₯) = 0 24/5βπ₯/50 = 0Putting Pβ(π) = 0 24/5βπ₯/50 = 0 (βπ₯)/50=(β24)/5 π₯ = (β24)/5 Γ β50 π₯ = 240 Finding Pββ(π) Pβ(π₯) = 24/5 β π₯/50 Diff w.r.t π₯ Pββ(π₯) = π(24/5 β π₯/50)/ππ₯ = 0 β 1/50 = (β1)/50 < 0 Since Pββ(π) < 0 at π₯ = 240 β΄ π₯ = 240 is point of maxima Thus, P(π₯) is maximum when π = 240 Hence the manufacturer can earn maximum profit if he sells 240 items.