Expert: Paul Klarreich - 9/13/2007

Question
Hi, Paul

Could you please help the solve the following question as it took me days to figure out the answer and I end up with nothing........please help and please be quick as my assignment is due tomorrow. The question is:

The per period sales of a new product, x(t), evolves over time according to

x(t):=  A / (1 + b * e ^ – c t )  where A, b and c are positive constants.

a) By taking the limit of x(t) as t tends to infinity, show that per period sales tends to A as t increases.

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

c) Show that the maximum rate of growth occurs when x(t) := A / 2

d) Find the value of t, as a function of the constants b and c, at which this maximum growth occurs.

Thank you,
Camilla  

Answer
Questioner:   camilla
Category:  Advanced Math
Private:  No
 
Subject:  Calculus
Question:  Hi, Paul

Could you please help the solve the following question as it took me days to figure out the answer and I end up with nothing........please help and please be quick as my assignment is due tomorrow. The question is:

The per period sales of a new product, x(t), evolves over time according to

x(t):=  A / (1 + b * e^-ct )  where A, b and c are positive constants.

a) By taking the limit of x(t) as t tends to infinity, show that per period sales tends to A as t increases.

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

c) Show that the maximum rate of growth occurs when x(t) := A / 2

d) Find the value of t, as a function of the constants b and c, at which this maximum growth occurs.

Thank you,
Camilla
.......................................

Hi, Camilla,

One of my rules for questioners is:

I cannot answer questions that use special jargon from engineering or business applications.

because I have no idea what a 'per period sales' means.  Another reason for this rule is that if you take care to eliminate this jargon and express your answer in basic mathematical terms, you might find that the difficulty disappears.  So I will try to restate your problem in basic math terms:

The per period sales of a new product, x(t), evolves over time according to

If  x(t) =  A/(1 + b * e^-ct )  where A, b and c are positive constants.

a) Show that the limit of x(t) as t -> infinity = A.

b) Show that  dx/dt = K(A - x), where K is a constant.

c) Show that the maximum rate of growth occurs when  x(t) = A / 2.  (This one is good.)

d) Find the value of t, at which this maximum growth occurs.

OK, now that we are just doing math, we can get started:

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

a) Show that the limit of x(t) as t -> infinity = A.

It is easy to show that  e^-n approaches zero as  n -> infinity.

So  lim[t->inf] e^-ct = 0, since (you said) c is a positive constant.  Then:
             A
lim     ------------ =
t->inf  1 + b e^-ct

   A
-------- =
1 + b(0)

   A
-------- =  A
   1

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

b) Show that the dx/dt = K(A - x), where K is a constant.

THIS ONE DOES NOT SEEM TO WORK OUT.  Perhaps something was left out, or I am missing something.  It seems to go like this:

Find  dx/dt using the chain rule:

        A
x =  ----------- = A(1 + b e^-ct)^-1
    1 + b e^-ct

Use the chain rule:  x = Au^-1,  where  u = 1 + b e^-ct
dx
-- = -Au^-2,  and
du

du
-- = -bc e^-ct
dt

dx
--  =  -A(1 + b e^-ct)^-2(-bc e^-ct)
dt

dx       Abc e^-ct
--  =  ---------------
dt     (1 + b e^-ct)^2

Now that's your rate of change, and you have to show that it is  K(A - x)
                A
A - x = A - ------------- =
           (1 + b e^-ct)  

A(1 + b e^-ct) - A
------------------- =
  (1 + b e^-ct)  

  Ab e^-ct
--------------- =
(1 + b e^-ct)  

Abc e^-ct(1 + b e^-ct)
----------------------- =
 c(1 + b e^-ct)^2

dx (1 + b e^-ct)
-- ---------------- =
dt       c

dx A(1 + b e^-ct)
-- --------------- =
dt       Ac

dx  A
-- ---
dt  xc

So what I have here is:

dx  A
-- --- = A - x
dt  xc

dx  A
-- --- = (A - x)xc/A
dt  xc

or that

dx
-- = Kx(A - x), where  K = c/A
dt
.................................
c) Show that the maximum rate of growth occurs when  x(t) = A / 2.  (This one is good.)

That means dx/dt has a maximum for some value of t.  So we will want the second derivative:

dx       Abc e^-ct
--  =  ---------------
dt     (1 + b e^-ct)^2

We can ignore the Abc part for this:

dx         Abc e^-ct
--  ~~  ---------------
dt      (1 + b e^-ct)^2

We need the quotient rule:

     ((1 + b e^-ct)^2)(-c e^-ct) - (e^-ct)(2(1 + b e^-ct)(-bc e^-ct)
x'' = ----------------------------------------------------------------
         ()^4

     (1 + b e^-ct)(-c e^-ct) - (e^-ct)(2)(-bc e^-ct)
x'' = ---------------------------------------------------
                  ()^3

     e^-ct[(1 + b e^-ct)(-c) - (e^-ct)(2)(-bc)]
x'' = -------------------------------------------
                  ()^3

     e^-ct[-c  - bc e^-ct + 2bc(e^-ct)]
x'' = -------------------------------------------
                  ()^3

     c e^-ct[-1 - b e^-ct + 2b e^-ct]
x'' = -------------------------------------------
                  ()^3
 
     c e^-ct[-1 + b e^-ct]
x'' = ----------------------
              ()^3


OK, now that will be zero (getting your maximum) when -1 + b e^-ct = 0

or  when    b e^-ct = 1

Whenever that is, our original function:

        A
x =  -----------
    1 + b e^-ct

will be equal to:

      A
x =  ----- = A/2
    1 + 1
..................................
d) Find the value of t, at which this maximum growth occurs.

After all that stuff, this one is easy:

b e^-ct = 1

e^-ct = 1/b

-ct = ln(1/b) = - ln b

t = (ln b)/c


Whew!