Expert: Paul Klarreich - 9/24/2009

Question
Hi Paul, just wondering if you could help me out with this question.  Hopefully it wont take up alot of your time as i understand you get many questions a day.



Given the equation of a per period sales of a product "x(t)"

x(t) = A/(1+be^-ct)


Where "A" "b" "c" are positive constants and  "e" is exponential

(a) Show that rate of change of sales is proportional to the difference between A and x(t)

(b) Show that maximum rate of change occurs at x(t) = A/2

(c) find value of t as a function of the constants b and c, at which the above maximum rate occurs at which this maximum change occurs


Thanks in advance Paul!

Sincerely

Greg

Answer
Questioner: Greg
Country: Australia
Category: Advanced Math
Private: No
Subject: finding rate of change.
Question: Hi Paul, just wondering if you could help me out with this question.  Hopefully it wont take up alot of your time as i understand you get many questions a day.

Given the equation of a per period sales of a product "x(t)"

x(t) = A/(1+be^-ct)

>> OK.
Where "A" "b" "c" are positive constants and  "e" is exponential

>> OK.

(a) Show that rate of change of sales is proportional to the difference between A and x(t)

>> OK.  rate of change of sales means  x'(t).
>> 'is proportional to' means  = K times it.
>> 'the difference between A and x(t)' means  A - x(t), which in this case means:
A - A/(1+be^-ct)

Find  x'(t)

x(t) = A/(1+be^-ct) = A/u, where  u = 1 + be^-ct
dx/du = -A/u^2,   du/dt = - bc e^-ct
        Abc e^-ct
x'(t) =  ------------
       (1+be^-ct)^2
..................................
Now A - x(t) =
      A
A - ---------- =
   (1+be^-ct)

A(1 + be^-ct) - A
-----------------
  (1+be^-ct)

A + Abe^-ct - A
----------------
  1+be^-ct

 Abe^-ct
-----------
1+be^-ct

This does not look right.  In fact, if we take  A = 1, b = 1,  c = 1, for example, and look at:
exp(-x)
         e^-t   exp(-x)
x'(t) =  ---------
       (1+e^-t)^2
.........................and.............
A - x(t) =
      1
1 - -------- =
   (1+e^-t)

(1+e^-t) - 1
------------
(1+e^-t)


e^-t
--------
1+e^-t


These won't be the same (or even proportional).  Something is wrong.

(b) Show that maximum rate of change occurs at x(t) = A/2

That means you want the max of  x'(t):
        Abc e^-ct
x'(t) =  ------------
       (1+be^-ct)^2
         (1 + be^-ct)^2(-ce^-ct) - (e^-ct)2(1 + be^-ct)(-bc e^-ct)
x'' = Abc[-------------------------------------------------------- ]
             (1 + be^-ct)^4

         (1 + be^-ct)(-ce^-ct) - (e^-ct)2(-bc e^-ct)
x'' = Abc[-------------------------------------------------------- ]
             (1 + be^-ct)^3

         (1 + be^-ct)(-ce^-ct) - (e^-ct)2(-bc e^-ct)
x'' = Abc[------------------------------------------- ]
             (1 + be^-ct)^3


Call y = e^-ct:  (which will always be >0 )

         (1 + by)(-cy) - (y)2(-bc y)
x'' = Abc[---------------------------]
             (1 + by)^3

         -cy - bcy^2 + 2bc y^2
x'' = Abc[----------------------]
             (1 + by)^3

Set   -cy - bcy^2 + 2bc y^2 = 0

-c  + bc y = 0

   bc y = c

y = 1/b, or   e^-ct = 1/b, or e^ct = b,  or ct = ln b,  t = (1/c) ln b

Now if

x(t) = A/(1+be^-ct) = A/(1 + by) = A/(1 + b(1/b)) = A/(1 + 1) = A/2
.....................

(c) find value of t as a function of the constants b and c, at which the above maximum rate occurs at which this maximum change occurs

>> See above.

Thanks in advance Paul!