Ave number of cases over six months

[KLS2111]

A store orders 288 cases every 6 months. The number of cases remaining t months after the order arrives is f(t) = 288 e^(-.05t)

Find the average number of cases in inventory over the six month period.


I know this seems like an easy question.. and I am probably over thinking. If I take the antideriv or the deriv I beleive my answer is negative. So I took plugged 0 in and 6 in added my answers and then divided by 7. I have also found each month added them and then divided by 7. Of course I get different answers each time. So anyone who can point me in teh right direction would be greatly appreciated.

 

[stapel]

I get different answers each time. So anyone who can point me in teh right direction would be greatly appreciated.

Without seeing your work, I'm afraid it will be difficult to diagnose any errors or correct your direction.

Please reply showing all of your steps and reasoning. Thank you.

Eliz.

 

[KLS2111]

As I stated I tried taking the deriv and the anti deriv but I come up with a negative number of cases each time when I do that.

So then I plugged in each number 0-7 into f(t) and came up with the following data: 0 months 288, 1 month 174, 2 months 106, 3 months 64, 4 months 39, 5 months 24, 6 months 14. So then I added them up and divided by 7 and got 101 cases.

I also tried adding 288 and 14 ( 0 and 6 months) then dividing by 7 to get 43 cases.

This problem has multiple choice
95 91 85 546 48

I relaize that 548 is not possible because we wouldn't have more cases than what we started with. I hope this will help you point me in teh right direction.

 

[stapel]

As I stated I tried taking the deriv and the anti deriv but I come up with a negative number of cases each time when I do that.

Okay. But what were your steps? (We can't diagnose a description of a method and a result. We really do need to see what you've done.)

So then I plugged in each number 0-7 into f(t)...then I added them up and divided by 7 and got 101 cases.

This process is an approximation of the antiderivative (though I'm not sure where "7" is coming from; the exercise specifies "six" month). Since you did not get a negative answer for your approximation, this would suggest that there was some error with your antiderivative.

Note: Your function values cannot be derived from the function you posted. Are you sure the power is "-0.05t" and not "-0.5t"?

Thank you.

Eliz.

 

[KLS2111]

Ok maybe this will help

You are correct... the problem should read 288e^(-.5t)

When I took the antideriv I came up with -576 e^(-.5t) +c

After zero months passed I there were 288 cases right?

So I did 288 = -576e^(0) +C
C = 864

So the then I need to plug 0 and 6 into -576e^(-.5t)+ 864

Then I get 835 - 288= 547

How can there be more cases as the average than what we started with?

 

[stapel]

Re: Ok maybe this will help

You are correct... the problem should read 288e^(-.5t)

Thank you!

When I took the antideriv I came up with -576 e^(-.5t) +c

Now evaluate at t = 6 and at t = 0. Then divide by 6 (the interval length) to get the average value of the function. You should get one of the listed options as your (rounded) answer.

Eliz.