course Mth 271
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RESPONSE -->
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18:34:11 2.3.32 P=22t^2+52t+10000, t from 1970; find P at t=0,10,20,25 and explain; find dP/dt; evaluate at given t and explain your results.
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RESPONSE --> P(0) = 10,000 P(10) = 12,720 P(20) = 19,840 P(25) = 25,050 P' = 44t + 52 P'(0) = 52 P'(10) = 492 P'(20) = 932 P'(25) = 1152 confidence assessment: 3
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18:34:14 dP/dt=44t + 52 (power function rule on each nonconstant term) When t = 0, 10, 20 and 25 you would have P = 10,000, 12,700, 20,000, 25,000 approx. At these values of t we have dP / dt = 52, 492, 932 and 1152 (these are my mental calculations--check them). dP / dt is the rate of change of the population with respect to time t **
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RESPONSE --> self critique assessment: 3
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18:34:25 2.3.48 demand fn p = 50/`sqrt(x), cost .5x+500. Find marginal profit for x=900,1600,2500,3600 Explain how you found the marginal profit, and give your results.
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RESPONSE --> Profit = Revenue - cost Revenue = 50 sqrt(x) Cost = .5x + 500 P = 50 sqrt(x) - .5x + 500 P'(x) = 25/sqrt(x) - .5 P'(900) = .33 P'(1600) = .13 P'(2500) = 0 P'(3600) = -.08 confidence assessment: 3
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18:34:27 x represents the number of items sold. If x items are sold at price p = 50 / `sqrt(x), then revenue is price of item * number sold = 50 / `sqrt(x) * x = 50 `sqrt(x). The profit is revenue - cost = 50 `sqrt(x) - .5 x - 500. The marginal profit is the derivative of the profit function, which is (50 `sqrt(x) - .5 x - 500 ) ' = 25 / `sqrt(x) - .5. Evaluating the marginal profit at x = 900, 1600, 2500 and 3600 we get values .33..., .125, 0 and -.0833... . This shows us that the marginal profit, which is the limiting value of the increase in profit per additional item manufactured, is positive until x = 2500. This means that it is to the advantage of the producer to produce new items when x = 900 and when x = 1600, but that the advantage disappears as soon as x reaches 2500. So 2500 is the best selling price. When x = 3600 production of additional items reduces profits. **
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RESPONSE --> self critique assessment: 3
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