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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part 2 - Question 2 - Case Based Questions (MCQ) - Serial order wise - Chapter 6 Class 12 Application of Derivatives

Question 1

What will be the production when the profit is maximum?

(a) 37500

(b) 12.5

(c) –12.5

(d) –37500

part 3 - Question 2 - Case Based Questions (MCQ) - Serial order wise - Chapter 6 Class 12 Application of Derivatives

part 4 - Question 2 - Case Based Questions (MCQ) - Serial order wise - Chapter 6 Class 12 Application of Derivatives
part 5 - Question 2 - Case Based Questions (MCQ) - Serial order wise - Chapter 6 Class 12 Application of Derivatives

Ā 

Question 2

What will be the maximum profit?

(a) Rs 38,28,125

(b) Rs 38281.25

(c) Rs 39,000

(d) None

part 6 - Question 2 - Case Based Questions (MCQ) - Serial order wise - Chapter 6 Class 12 Application of Derivatives

Ā 

Question 3

Check in which interval the profit is strictly increasing .

(a) (12.5, āˆž)

(b) for all real numbers

(c) for all positive real numbers

(d) (0, 12.5)

part 7 - Question 2 - Case Based Questions (MCQ) - Serial order wise - Chapter 6 Class 12 Application of Derivatives

Ā 

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

part 8 - Question 2 - Case Based Questions (MCQ) - Serial order wise - Chapter 6 Class 12 Application of Derivatives

Ā 

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

part 9 - Question 2 - Case Based Questions (MCQ) - Serial order wise - Chapter 6 Class 12 Application of Derivatives part 10 - Question 2 - Case Based Questions (MCQ) - Serial order wise - Chapter 6 Class 12 Application of Derivatives

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Transcript

Question P(x) = ć€–āˆ’5š‘„ć€—^2+125š‘„+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) –12.5 (d) –37500 Given P(x) = āˆ’5x2 + 125x + 37500 Differentiating w.r.t x P’(x) = d(ć€–āˆ’5š‘„ć€—^2 + 125š‘„ + 37500)/š‘‘š‘„ P’(x) = āˆ’5 Ɨ 2x + 125 P’(x) = āˆ’10x + 125 Putting P’(x) = 0 āˆ’10x + 125 = 0 āˆ’10x = --125 x = (āˆ’125)/(āˆ’10) x = 12.5 Finding P’’(x) P’’(x) = d(āˆ’10š‘„ + 125)/š‘‘š‘„ P’’(x) = āˆ’10 < 0 Since P’’(š’™) < 0 at š‘„ = 12.5 ∓ š‘„ = 12.5 is point of maxima Thus, P(š‘„) is maximum when š’™ = 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) = āˆ’5x2 + 125x + 37500 = āˆ’5(12.5)2 + 125(12.5) + 37500 = āˆ’5(156.25) + 1562.5 + 37500 = āˆ’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, āˆž) (b) for all real numbers (c) for all positive real numbers (d) (0, 12.5) Profit is strictly increasing where P’(x) > 0 āˆ’10x + 125 > 0 125 > 10x 10x < 125 x < 12.5 So, profit is strictly increasing for x ∈ (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) = āˆ’5x2 + 125x + 37500 = āˆ’5(2)2 + 125(2) + 37500 = āˆ’5(4) + 250 + 37500 = āˆ’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) = āˆ’5x2 + 125x + 37500 38250 = āˆ’5x2 + 125x + 37500 5x2 āˆ’ 125x āˆ’ 37500 + 38250 = 0 5x2 āˆ’ 125x + 750 = 0 Dividing both sides by 5 x2 āˆ’ 25x + 150 = 0 x2 āˆ’ 10x āˆ’ 15x + 150 = 0 x(x āˆ’ 10) āˆ’ 15(x āˆ’ 10) = 0 (x āˆ’ 10) (x āˆ’ 15) = 0 ∓ x = 10, 15 Since x = 15 is in the options So, the correct answer is (a)

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo