Expert: Paul Klarreich - 11/22/2005

Question
1-A company determines that it can sell 1000 units of a product per month if the price is 5 birr for each unit. it also estimates that for each one cent reduction in unit price 10 more units can be sold. under these conditions,what is the maximum possible income and what price per unit gives this income?
2-show that:
f(x)=x*x*x+4/x*x+7 has only one zero on interval on an open interval negative infinity and 0.
3-Use L'hopital's rule to evaluate the following limits if the exist
a.  
lim x/ln(1+2ex)
  x gose to positive infinity
b.  lim [1/lnx-1/x-1]
    x gose to 1 from the right
c.  lim sin(3+x)(3+x)-sin9/x
      x gose to 0
d.                2x+1  
  lim (2x-3/2x+5)
x gose to positive infinity  

Answer
Hi, Dessalegn,

I must ask a couple of things from you in the future:
(A) Don't send duplicate questions.  There is a limit on the number of questions that can be sent to me in a day, and you should not use up two of those with the same question.  It is good, of course, that you combined several problems into one question.
(B) When you send a question, let me know what you tried to do, already.  I might be able to tell where you went wrong and that may be more valuable to you than just the answer.

1-A company determines that it can sell 1000 units of a product per month if the price is 5 birr (DOLLARS?) for each unit. it also estimates that for each one cent reduction in unit price 10 more units can be sold. under these conditions,what is the maximum possible income and what price per unit gives this income?


Choose your variables:

Let p = price per unit.
   s = sales per month.
   I = total income.

Write your functions:
If the price is p = 5, s = 1000
For each 'difference in price' of 0.01, add 10.  Then for each 'difference in price' of 1.00, add 1000 --makes it easier.

Then
s = 1000 + 1000(5 - P)

and

I = Ps = P(1000 + 1000(5 - P))

You're on your way now.  Just simplify, differentiate (get dI/ds) and set the derivative equal to zero, and solve.



2-show that:
f(x)=x*x*x+4/x*x+7 has only one zero on interval on an open interval negative infinity and 0.

Approach:  This function is continuous and differentiable on the open interval (-inf,0).  Suppose that f(x1) = 0 AND f(x2) = 0, where x1 and x2 are both in that interval.  Then you can use Rolle's Theorem, which says that there must be a point x0, between x1 and x2 where f'(x0) = 0.  Try to find such a point:

f'(x) = 3x^2 - 8/x^3

Set that equal to zero and simplify.  You will get:

3x^5 = 8
x = fifth root of (8/3), which will be a positive number.  (Actually there are complex roots, too, but they don't concern us.)

So this derivative is never zero on (-inf,0), therefore there cannot be two zeroes of the original function.



3-Use L'hopital's rule to evaluate the following limits if the exist
a.
lim x/ln(1+2ex)
  x --> + infinity
b. lim [1/lnx-1/x-1]
   x goes to 1 from the right
c. lim sin(3+x)(3+x)-sin9/x
      x goes to 0
d.                2x+1
  lim (2x-3/2x+5)
x goes to positive infinity


a.
lim    x/ln(1+2e^x)   [exp(x) means  e^x]
x->+inf

Differentiate top and bottom:
        1              1 + 2exp(x)
------------------   =  ------------
   1                     2exp(x)
-----------(2exp(x))
1 + 2exp(x)

Now you can do either of two things:
(1) Repeat the process and get 2exp(x)/2exp(x) which -> 1
(2) Multiply top and bottom by exp(-x) and get
exp(-x) + 2           0 + 2
------------ which -> ------ = 1
     2                 2



b. lim [1/lnx-1/x-1]
   x gose to 1 from the right

Your general scheme when you have an indeterminate for like (inf - inf), for which you CANNOT use l'Hospital's rule (note: small l, big H, s after o), is to do some
algebra to make it  0/0 or inf/inf.  In this case, try combining the fractions

over an LCD:
 1       1      x - 1 - ln x  
----- - ------ = ------------ -> 0/0;  good!
ln x    x - 1    ln x(x - 1)   

Now use your rule, and differentiate.  You will need the product rule.
  1 - 1/x                1 - 1/x
-------------------- = ---------------
ln x(1) + 1/x(x - 1)   ln x + 1 - 1/x

Multiply through by x:
   x - 1
--------------  still -> 0/0, so repeat the process.
x ln x + x - 1
      1          1
------------ -->  -
1 + ln x + 1      2


c. lim sin(3+x)(3+x)-sin9/x
      x gose to 0

THIS ONE MAKES NO SENSE AS WRITTEN.  Perhaps you mean:
       sin(3+x)(3+x)-sin9      
lim   --------------------
x->0            x

Please resubmit.


d.                
  lim ((2x-3)/(2x+5))^(2x+1)
x gose to positive infinity

Try to make this look something like  (1 + 1/n)^n and then we can use the logarithmic form of l'Hospital's rule.

Let n = 2x+1;  then 2x - 3 = n - 4, and 2x + 5 = n + 4, so you have:

((n - 4)/(n + 4))^n

Now the log of this is:

n [ln(n - 4) - ln(n + 4)] = [ln(n - 4) - ln(n + 4)]/(1/n)

(It has to be a fraction, remember?)

Now apply the rule:
 1       1
----- - -----
n - 4   n + 4
-------------
    -1
    ---
    n^2

Multiply out, using n^2(n-4)(n+4) as your LCM.

n^2(n + 4) - n^2(n - 4)   4n^2 + 4n^2      8n^2
----------------------- = ----------- = ----------  --> -8, when n->0
   - (n-4)(n+4)          -(n^2 - 16)   -n^2 + 16

That's it, mostly.  Since ln(....) = -8, the answer is e^(-8)