Assignment 15 query

course Mth 271

015. `query 15*********************************************

Question: `q 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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

P = 22t^2 + 52t + 10000

‘dP/dt = 44t + 52

t ‘dP/dt

0 52

10 492

20 932

25 1152

P = 22t^2 + 52t + 10000

t P

0 10000

10 12720

20 19840

25 25050

The derivative is the rate of change.

confidence rating:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

I feel very confident this problem.

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

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 Rating:

I took the derivative of the function, evaluated t, then evaluated t in the original function.

*********************************************

Question: `q 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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

revenue = p = [50/’sqrt(x)](x) = 50’sqrt(x)

cost = .5x + 500

profit = revenue – cost

profit = P = (50’sqrt) - .5x – 500

marginal profit = P’ = 25x^(-1/2) - .5

x P

900 .33

1600 .125

2500 0

3600 -.083

confidence rating:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

I feel confident about my answer.

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

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. **

------------------------------------------------

Self-critique Rating:

I used logic to find the revenue and the cost. I took the derivative of the sum of these two values and used that value as the marginal profit. From here I plugged in the values of x to get the profit.

"

xxxx

&#Good responses. Let me know if you have questions. &#