open qa 152

#$&*

course Mth 271

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: t = 0 P = 10 000 T = 10 P = 22 * 100 + 52 * 10 + 10 000 = 12 720 T = 20 P = 19840 T = 25 P = 25050 Dp/dt = 44t + 52 Dp/dt (10) = 492 Dp/dt (0) = 52 dp/dt (20) = 939 dp/dt (25) = 1152 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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): ------------------------------------------------ Self-critique Rating:

********************************************* 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: R = xp = 50x/sqrt(x) P = R - C P = 50x/sqrt(x) - 0.5x - 500 Marginal profit is the derivative of the profit function. dP/dx = [50 sqrt(x) - 1/2sqrt(x) * 50x]/x - 0.5 = [50sqrt(x) - 25x/sqrt(x)]/x - 0.5 = 25x/x*sqrt(x) - 0.5 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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 didn’t complete the plugging in part because I seem to have one X extra in my derivative and I am not sure where I messes up…. That is I see where I messed up but I am still not sure why is x * 50/sqrt(x) = 50 * sqrt(x) and not 50x/sqrt(x)…. I am sure this is something I SHOULD know… however it doesn’t click right now…. ------------------------------------------------ Self-critique Rating:ok

open qa 152

#$&*

course Mth 271

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:

t = 0

P = 10 000

T = 10

P = 22 * 100 + 52 * 10 + 10 000 = 12 720

T = 20

P = 19840

T = 25

P = 25050

Dp/dt = 44t + 52

Dp/dt (10) = 492

Dp/dt (0) = 52

dp/dt (20) = 939

dp/dt (25) = 1152

confidence rating #$&*:

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

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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 **

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Self-critique (if necessary):

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

Self-critique Rating:

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

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:

R = xp = 50x/sqrt(x)

P = R - C

P = 50x/sqrt(x) - 0.5x - 500

Marginal profit is the derivative of the profit function.

dP/dx = [50 sqrt(x) - 1/2sqrt(x) * 50x]/x - 0.5 = [50sqrt(x) - 25x/sqrt(x)]/x - 0.5 = 25x/x*sqrt(x) - 0.5

confidence rating #$&*:

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

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

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)).

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Self-critique (if necessary):

I didn’t complete the plugging in part because I seem to have one X extra in my derivative and I am not sure where I messes up…. That is I see where I messed up but I am still not sure why is x * 50/sqrt(x) = 50 * sqrt(x) and not 50x/sqrt(x)…. I am sure this is something I SHOULD know… however it doesn’t click right now….

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Self-critique Rating:ok