query 15

course mth 271

è²æªLãÙÀ•Æ“l¬|îúGó™Ú½åyÍ꯯assignment #015

015. `query 15

Applied Calculus I

11-06-2007

......!!!!!!!!...................................

10:30:27

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.

......!!!!!!!!...................................

RESPONSE -->

this problem wasnt assigned but...

(0,10000) (10,12720) (15,15730) (20,19840) (25,25056)

p'=44t+52

(0,52) (10,492) (15,712) (20,932) (25,1152)

p' is the rate of change of pop wrt time t

confidence assessment: 3

.................................................

......!!!!!!!!...................................

10:30:38

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 **

......!!!!!!!!...................................

RESPONSE -->

yup

self critique assessment: 3

.................................................

......!!!!!!!!...................................

10:35:22

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.

......!!!!!!!!...................................

RESPONSE -->

how do you get the deriv of 50/sq root x?

confidence assessment: 0

.................................................

......!!!!!!!!...................................

10:40:27

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. **

......!!!!!!!!...................................

RESPONSE -->

i wish you worked out all of the steps in finding the deriv b/c sometimes it is hard to follow how you are getting the answers

self critique assessment: 3

The derivative of the square root has been giving you some trouble; see my notes on preceding queries and qa's and let me know if the given solution doesn't then become clear.

If there are too many details in the solution then the reader tends to 'lose for forest for the trees'. The point of this problem is to apply derivatives to this situation; it is assumed that at this point you know how to take the derivative of the basic functions. Of course that isn't always the case, which is why I encourage you to ask.

.................................................

&#

Your work looks good. Let me know if you have any questions. &#