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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Transcript

Question P(x) = 怖āˆ’5š‘„怗^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) ā€“12.5 (d) ā€“37500 Given P(x) = āˆ’5x2 + 125x + 37500 Differentiating w.r.t x Pā€™(x) = d(怖āˆ’5š‘„怗^2 + 125š‘„ + 37500)/š‘‘š‘„ Pā€™(x) = āˆ’5 Ɨ 2x + 125 Pā€™(x) = āˆ’10x + 125 Putting Pā€™(x) = 0 āˆ’10x + 125 = 0 āˆ’10x = --125 x = (āˆ’125)/(āˆ’10) x = 12.5 Finding Pā€™ā€™(x) Pā€™ā€™(x) = d(āˆ’10š‘„ + 125)/š‘‘š‘„ Pā€™ā€™(x) = āˆ’10 < 0 Since Pā€™ā€™(š’™) < 0 at š‘„ = 12.5 āˆ“ 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 Pā€™(x) > 0 āˆ’10x + 125 > 0 125 > 10x 10x < 125 x < 12.5 So, profit is strictly increasing for x āˆˆ (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 āˆ“ 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