Ex 6.1, 18 (MCQ) - Chapter 6 Class 12 Application of Derivatives

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Ex 6.1, 18 - Total revenue is given by R(x) = 3x^2 + 36x + 5. Marginal

Ex 6.1, 18 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.1, 18 - Chapter 6 Class 12 Application of Derivatives - Part 3

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Ex 6.1, 18 The total revenue in Rupees received from the sale of š‘„ units of a product is given by R(š‘„) = 3š‘„2 + 36š‘„ + 5. The marginal revenue, when š‘„ = 15 is (A) 116 (B) 96 (C) 90 (D) 126Marginal revenue is rate of change of total revenue w. r. t the number of unit sold Let MR be marginal revenue So, MR = š’…š‘¹/š’…š’™ Given, Total revenue = R (š‘„) = 3š‘„2 + 36š‘„ + 5 We need to find marginal revenue when š‘„ = 15 i.e. MR when š‘„ = 15 MR = š‘‘(š‘…(š‘„))/š‘‘š‘„ MR = (š‘‘ (3š‘„2 + 36š‘„ + 5) )/š‘‘š‘„ MR = (š‘‘(3š‘„2))/š‘‘š‘„ + (š‘‘(36š‘„))/š‘‘š‘„ + (š‘‘(5))/š‘‘š‘„ MR = 3 (š‘‘(š‘„2))/š‘‘š‘„ + 36 (š‘‘(š‘„))/š‘‘š‘„ + 0 MR = 3 Ɨ 2š‘„ + 36 MR = 6š’™ + 36 MR when š’™ = 15 MR = 6(15) + 36 MR = 90 + 36 MR = 126 Hence, the required marginal revenue is Rs. 126 Thus, D is the correct Answer

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