Example 37 - Chapter 6 Class 12 Application of Derivatives
Last updated at Dec. 16, 2024 by Teachoo
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Example 37 Manufacturer can sell š„ items at a price of rupees (5āš„/100) each. The cost price of š„ items is Rs (š„/5+500) Find the number of items he should sell to earn maximum profit.Let S(š) be the Selling Price of š„ items. & C(š) be the cost Price of š„ item. Given Manufacture sell š„ items at a price of rupees (5āš„/100) each S(š) = š„ Ć (5āš„/100) = 5š ā šš/ššš Also given Cost of x items is Rs. (š„/5+500) C(š„) = š„/5 + 500 We need to maximize profit Let P(š„) be the profit Profit = Selling price ā cost price P(š„) = S(š„) ā C(š„) P(š„) = (5š„āš„2/100)ā(š„/5+500) = 5š„ ā š„2/100ā š„/5 ā 500 = (25š„ ā š„)/5 ā š„2/100 ā500 = 24š„/5āš„2/100 "ā 500" Hence, P(š„) = 24š„/5āš„2/100ā500 Diff w.r.t x Pā(š„) = š(24/5 š„ ā š„^2/100 ā 500)/šš„ Pā(š„) = 24/5ā2š„/100ā0 Pā(š„) = 24/5āš„/50 Putting Pā(š„) = 0 24/5āš„/50 = 0 Putting Pā(š) = 0 24/5āš„/50 = 0 (āš„)/50=(ā24)/5 š„ = (ā24)/5 Ć ā50 š„ = 240 Finding Pāā(š) Pā(š„) = 24/5 ā š„/50 Diff w.r.t š„ Pāā(š„) = š(24/5 ā š„/50)/šš„ = 0 ā 1/50 = (ā1)/50 < 0 Since Pāā(š) < 0 at š„ = 240 ā“ š„ = 240 is point of maxima Thus, P(š„) is maximum when š = 240 Hence the manufacturer can earn maximum profit if he sells 240 items.
