#$&* course Mth 271 11/7 630p 5. `query 15
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Given Solution: `a 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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):Not profit, but population
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Given Solution: `a 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. ** STUDENT COMMENT I can see where I messed up. The only thing that I seemed to have gotten rigth is that Marginal Profit is the derivative of the Profit function. I still have problems taking derivatives with the square root function. INSTRUCTOR RESPONSE sqrt(x) can be written as x^(1/2). Using the fact that the derivative of x^n is n * x^(n-1) the derivative of x^(1/2) is 1/2 x^(1/2 - 1) = 1/2 x^(-1/2), or 1/2 * (1 / x^(1/2) ) = 1 / (2 x^(1/2)), alternatively 1 / (2 sqrt(x)). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I felt like that cost formula was there for a reason, but I didn’t include it. I went back and computed the derivative with the cost formula included, which amounted to subtracting -.5 from each number.