P(x) = -5x 2 +125x + 37500 is the total profit function of a
company, where x is the production of the company.

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Question 1

What will be the production when the profit is maximum?

(a) 37500

(b) 12.5

(c) –12.5

(d) –37500

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Question 2

What will be the maximum profit?

(a) Rs 38,28,125

(b) Rs 38281.25

(c) Rs 39,000

(d) None

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Question 3

Check in which interval the profit is strictly increasing .

(a) (12.5, ∞)

(b) for all real numbers

(c) for all positive real numbers

(d) (0, 12.5)

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Question 4

When the production is 2 units, what will be the profit of the company?

(a) 37,500

(b) 37,730

(c) 37,770

(d) None

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Question 5

What will be production of the company when the profit is Rs 38,250?

(a) 15

(b) 30

(c) 2

(d) data is not sufficient to find

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  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

Transcript

Question P(x) = ใ€–โˆ’5๐‘ฅใ€—^2+125๐‘ฅ+37500 is the total profit function of a company, where x is the production of the company. Question 1 What will be the production when the profit is maximum? (a) 37500 (b) 12.5 (c) โ€“12.5 (d) โ€“37500 Given P(x) = โˆ’5x2 + 125x + 37500 Differentiating w.r.t x Pโ€™(x) = d(ใ€–โˆ’5๐‘ฅใ€—^2 + 125๐‘ฅ + 37500)/๐‘‘๐‘ฅ Pโ€™(x) = โˆ’5 ร— 2x + 125 Pโ€™(x) = โˆ’10x + 125 Putting Pโ€™(x) = 0 โˆ’10x + 125 = 0 โˆ’10x = --125 x = (โˆ’125)/(โˆ’10) x = 12.5 Finding Pโ€™โ€™(x) Pโ€™โ€™(x) = d(โˆ’10๐‘ฅ + 125)/๐‘‘๐‘ฅ Pโ€™โ€™(x) = โˆ’10 < 0 Since Pโ€™โ€™(๐’™) < 0 at ๐‘ฅ = 12.5 โˆด ๐‘ฅ = 12.5 is point of maxima Thus, P(๐‘ฅ) is maximum when ๐’™ = 12.5 So, the correct answer is (b) Question 2 What will be the maximum profit? (a) Rs 38,28,125 (b) Rs 38281.25 (c) Rs 39,000 (d) None Putting x = 12.5 in P(x) P(x) = โˆ’5x2 + 125x + 37500 = โˆ’5(12.5)2 + 125(12.5) + 37500 = โˆ’5(156.25) + 1562.5 + 37500 = โˆ’781.25 + 1562.5 + 37500 = 781.25 + 37500 = Rs 38281.25 So, the correct answer is (b) Question 3 Check in which interval the profit is strictly increasing . (a) (12.5, โˆž) (b) for all real numbers (c) for all positive real numbers (d) (0, 12.5) Profit is strictly increasing where Pโ€™(x) > 0 โˆ’10x + 125 > 0 125 > 10x 10x < 125 x < 12.5 So, profit is strictly increasing for x โˆˆ (0, 12.5) So, the correct answer is (d) Question 4 When the production is 2 units, what will be the profit of the company? (a) 37,500 (b) 37,730 (c) 37,770 (d) None Putting x = 2 in P(x) P(x) = โˆ’5x2 + 125x + 37500 = โˆ’5(2)2 + 125(2) + 37500 = โˆ’5(4) + 250 + 37500 = โˆ’20 + 250 + 37500 = 230 + 37500 = Rs 37730 So, the correct answer is (b) Question 5 What will be production of the company when the profit is Rs 38,250? (a) 15 (b) 30 (c) 2 (d) data is not sufficient to find Putting P(x) = 38250 in formula of P(x) P(x) = โˆ’5x2 + 125x + 37500 38250 = โˆ’5x2 + 125x + 37500 5x2 โˆ’ 125x โˆ’ 37500 + 38250 = 0 5x2 โˆ’ 125x + 750 = 0 Dividing both sides by 5 x2 โˆ’ 25x + 150 = 0 x2 โˆ’ 10x โˆ’ 15x + 150 = 0 x(x โˆ’ 10) โˆ’ 15(x โˆ’ 10) = 0 (x โˆ’ 10) (x โˆ’ 15) = 0 โˆด x = 10, 15 Since x = 15 is in the options So, the correct answer is (a)

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.