Another Word Problem :(

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hopelynnwelch

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I am doing great with derivatives. I have them down good. But I see a word problem and I have no idea how to set this problem up. These are the death of me. If I just knew how to set these up I could do them but I'm so clueless... ARG

Suppose the annual cost per active-duty armed service member in a certain country increased from $80,000 in 1995 to $90,000 in 2000. In 1990, there were 2 million armed service personnel and this number decreased to 1.5 million in 2000. Use linear models for annual cost and personnel to estimate, to the nearest $10 million, the rate of change of total military personnel costs in 1994.

Any help or hint appreciated...
 
hopelynnwelch said:
I am doing great with derivatives. I have them down good. But I see a word problem and I have no idea how to set this problem up. These are the death of me. If I just knew how to set these up I could do them but I'm so clueless... ARG

Suppose the annual cost per active-duty armed service member in a certain country increased from $80,000 in 1995 to $90,000 in 2000. In 1990, there were 2 million armed service personnel and this number decreased to 1.5 million in 2000. Use linear models for annual cost and personnel to estimate, to the nearest $10 million, the rate of change of total military personnel costs in 1994.

Any help or hint appreciated...
Click to expand...
Let 1995 mean t=0. So we have the point (80,000, 0)
So for year 2000, t=5. So the 2nd point is (90,000, 5).
What is the formula for the average rate of change? This was the hint you asked for.
Show us where you can go from here.
 
They want the rate of change in 1994. Don't they want a derivative and not an average?

I tried creating linear formulas and then multiplying them and taking the derivative of that. Then plugging 1994 into the derivative function... That is showing incorrect.

c(x)=.0005x-40
q(x)=-.00002x+40

F'(c(x) * q(x)) is -.0000002x+.0208

Then plugged in 1994. No this is not the answer, I tried.


For ARC how you stated Jomo

(90000-80000)/5-0 = 2000

I'm not sure where ARC helps me here but probably because I have no freaking clue what the heck I'm doing. I went to tutor and he wasted an hour and a half of my time to tell me he couldn't figure it out. So I have spent about 4 hours of my life on this silly problem.
 
Last edited:
hopelynnwelch said:
They want the rate of change in 1994. Don't they want a derivative and not an average?

I tried creating linear formulas and then multiplying them and taking the derivative of that. Then plugging 1994 into the derivative function... That is showing incorrect.

c(x)=.0005x-40 are you saying that c(2000)=90,000?
q(x)=-.00002x+40 are you saying that q(1990) = 1.5 million?

F'(c(x) * q(x)) is -.0000002x+.0208

Then plugged in 1994. No this is not the answer, I tried.


For ARC how you stated Jomo

(90000-80000)/5-0 = 2000

I'm not sure where ARC helps me here but probably because I have no freaking clue what the heck I'm doing. I went to tutor and he wasted an hour and a half of my time to tell me he couldn't figure it out. So I have spent about 4 hours of my life on this silly problem.
Click to expand...
Try getting the correct linear equations first
 
Okay.

Calling 1990 - 0

1995 - 5

and 2000 - 10

Points (80,000 , 5) and (90,000 , 10) For the first equation.

Points (2,000,000 , 0) and (1,500,000 , 10)

To get the slope, change in y over change in x

(10 - 5)/(90,000 - 80,000) = .0005 so m= .0005

(10 - 0)/(1,500,000 - 2,000,000) = -.00002 so m= -.00002

Then to find the equation I am using y-y1 = m(x-x1)

5=.0005(80,000)+b b = -35

0=-.00002(2,000,000)+b b = 40

Ohhh CRAP! LET ME TRY THIS! I hope I'm onto something here... lol

(.0005x - 35)(-.00002x + 40) = -.00000001x^2 + .02x + .0007x - 1400 = -.00000001x^2 +.0207x -1400

Now I am going to take the derivative of -.00000001x^2 +.0207x -1400

which is -.00000002x + .0207

Now I am going to plug 4 because I let 1990 be 0. So 1994 should be 4.

This gives me .02069992

That is not right.

A math tutor could not even save me. But I think I fixed the linear equation and I'm still wrong. I can't even move to the next section until this is right. So I've spent hours and hours and hours on it and I am going to fail this and everything after it because of this one stupid question.
 
I made an appointment with a different tutor tomorrow so hopefully that one is more help if no one can help figure this out by then so I won't have to fail all the next sections because I can't get this one problem right.
 
hopelynnwelch said:
I am doing great with derivatives. I have them down good. But I see a word problem and I have no idea how to set this problem up. These are the death of me. If I just knew how to set these up I could do them but I'm so clueless... ARG

Suppose the annual cost per active-duty armed service member in a certain country increased from $80,000 in 1995 to $90,000 in 2000. In 1990, there were 2 million armed service personnel and this number decreased to 1.5 million in 2000. Use linear models for annual cost and personnel to estimate, to the nearest $10 million, the rate of change of total military personnel costs in 1994.

Any help or hint appreciated...
Click to expand...
Sometimes words are used which are confusing and you have to learn them. An example from
http://www.purplemath.com/modules/translat.htm
"...gas was $4.12 a gallon..." mean if c is the cost of gas
c = 4.12 (dollars / gallon)
So if you got 2 gallons the price p was
p = 2 gallon * 4.12 (dollars / gallon) = 2 * 4.12 dollars = $8.24

So, turning words into formulas, we have to be general in defining our variable:
What do we want:
(1) The rate of change (over the years?) of total military personnel costs so p=total military personnel and c = costs [see (4) below]
(2) We want a value in 1994 [for x=6, see below]


What do we have
(3) they are throwing around years and we want the rate of change over the years so let y=year but why carry around all of those 1994, 1995, 2000, etc. Let's just carry around years since (smallest year) 1990 = x. So x = year-1994.
(4) they are throwing around costs and we want personnel costs, so let c = costs. However, there are a lot of zeros being thrown around so let the c be in thousands of dollars,
(5) x = 5 [y = 1995] then c = 80 [thousand dollars] in country A ('a certain country']
(6) x = 10 [y = 2000] then c = 90 [thousand dollars] in country A ('a certain country']
(7) x = 0 [y = 1990] then p = 2000 [thousand personnel, i.e. 2 million]
(8) x = 10 [y = 2000] then p = 1500 [thousand personnel, i.e. 1.5 million]
(9) Use linear models for annual cost and personnel.

Well linear is a line, so use (5) and (6) to get a line for c in terms of x and (7) and (8) to get a line for p in terms of x. We can use these lines to get an equation of costs in terms of personnel or costs in terms of elapsed years since 1990.

Now the rate of change in costs is the coefficient of p if the costs are expressed in number of personnel (change in costs per number of personnel) or the coefficient of x if the costs are expressed in x (change in costs per year). To put it another way, the rate of change of costs is the derivative of the costs.

Enough for now, got to go.
 
OMG!!! THANK YOU! THANK YOU! THANK YOU!


It is -300 and after all that help I finally did it!

After 7 hours on the same problem I can move on!


DID I SAY THANK YOU?

Thank you!
 
I guess I messed up what was the x value and what was the y value. I still don't really know how to think about these to figure that out but i finally found the answer to this question and so now I can finally sleep.
 
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