So Walt asks this question:

Mike, Timmy and Joe are 3 salesman for Assassin's Industries. So far, for the month of June Mike sold 2 times as much merchandise as Timmy, Joe sold (b-a) more than Mike, all together they sold a^2 and Joe does not like Timmy.

a. How much did they each sell?

b. If (a^2 - b) = sin^2(alpha) and a = cos^2(alpha), how much did they each and all sell? Does it make sense?

c. If part (b) does not make sense how about if a = 100 and b = 5000

d. If part (c) is less than optimal, will adding another salesperson, Lynda help? Knowing that optimal is maximized and Lynda sales anywhere from none to as much as Timmy sales?

e. Repeat part (c) and (d) with a = 1 and b = 10^3Challenge: Can you figure out what will be optimal if a = 50 and b = 100 if only 2 salespeople are to work? How about 3? Why?

Lets start with reading the keywords in the problem first, there are a few phrases that need our attention. Its important to spend a lot of time on reading the question over and over. You can read the problem once and take your chances, but unless your exceptionally good at translating a question correctly it is wise to reread. I thought I had understood the problem, and I did not reread. I am used to making a mistake and then back tracking until I find my error, its time consuming but I get the right answer eventually. The smart approach is to read carefully and then proceed with a clear understanding, you waste less time on a problem.

- [
**First clue**:] Mike sold 2 times as much merchandise as Timmy. What ever Timmy sold, Mike doubled. Or we can think of this as what ever Mike sold, Timmy sold only half as much. We will let 'm' represent Mike and 't' represent Timmy. We can denote this as so:

- [
**Second clue**:] Joe sold (b-a) more than Mike: Joe sold the same as Mike but with the addition of (b - a). We will let 'j' represent Joe. We are going to denote this as so:

- [
**Third clue**:] All together they sold:

- So if we are to combine the sales of Mike, Joe, and Timmy then our total sum should be equivalent to 'a' squared. We are going to denote this as so:

- [
**Fourth clue**:] Looking at all these formulas we see a common variable 'm', so we are going to denote our 'm' as so:

- [
**Part (a)**:] How much did they each sell? - Looking at each of the clues we can show the formula for each individual respectively. Lets start with Mike, his formula is simple because we are basing everyone's off of his sales.
- So we denote Mikes sales as such:

- Next, we look at Timmy's sales. We know from the clues that its half of Mike's sales.

- Now all we have left is Joe's sales, we are simply just pluging in what we initially stated for Mike's sales:

- [
**Part (b)**:] If (a^2 - b) = sin^2(alpha) and a = cos^2(alpha), how much did they each and all sell? Does it make sense? - In this part our three formulas for each individual can stay the same. Its the formula of how much they all sold together that has a significant change. The original formula looks like this:

- Now we can actually solve for our 'x' variable in terms of 'a' and 'b':

- With our given assumption that:

then

- But we don't stop here, there is a special relation with the sine function and the cosine function. If you look up trigonometric identities, you will see one of the Pythagorean Identities. So now your formula looks like this:

- So now we are saying that Mike's sale for June is 2/5, which is not really informative. This does not make much sense.

- [
**Part (c)**:] If part (b) does not make sense how about if a = 100 and b = 5000. - If we place actual values for 'a' and 'b' then we can see the following formula hold true:

- First lets solve for our variable 'x':

- Now we can check our initial formula for the total sum of sales for all three individuals being equal to 'a' squared:

- So now these assumptions check out, if a = 100 and b = 5000 then the conjecture holds true.

- [
**Part (d)**:] If part (c) is less than optimal, will adding another salesperson, Lynda help? Knowing that optimal is maximized and Lynda sales anywhere from none to as much as Timmy sales? - This one took some time to think about, what exactly is being asked of us here. We are keeping our values for 'a' and 'b' the same, lets look at the relationship between the variable 'x' and the total sum. We first have to calculate the value of 'x' then check if total sum is still equal to 'a' squared. Remember the original formulas from the beginning.

- solving for x we get

- Now the problem stated that Lynda can sell anywhere from as much as Timmy or less. So her sales can range from 0 to the value of Timmy's sales. We can denote her sales as so:

- I wrote a python code to start Lynda's sales close to 0 and then increased her sales through each iteration. You can slowly see how an increase in her sales makes a difference. Remember the bigger the denominator the closer to 0 the fraction gets. Her sales are really small in the beginning, but as her sales get bigger the total sales converges to 'a' squared. Look at the results so you can see the difference.

a = 100.0 b = 5000.0 i = 35 loop_condition = False while loop_condition == False: z = 1.0 + (1/2.0) + 1/(2.0**i) + 1.0 x = (1.0/z)*(a**2.0 + a - b) i -= 1 total_sales = (x + (x/2.0) + x + (b-a) + (x/2.0)) print 'the total sales equals: ', total_sales if (x + (x/2.0) + x + (b-a) + (x/2.0)) == a**2.0: print 'this is i', i loop_condition = True

- Lets assume she maximizes her sales, so her total sales value will be the same as Timmy's sale value:

- Then the following can be calculated:

- Now lets check to see if our total sum still is equal to 'a' squared:

- I just showed what the python code confirmed for us, only I did it by hand. So if Lynda sells as much as Timmy then the total sum of individuals is still equal to 'a' squared. We now know we can get the same results with only three people, so adding a fourth person does not help.

- [
**Part (e)**:] Repeat part (c) and (d) with a = 1 and b = 10^3.

- [
**Part (1) of (e)**:]

- So back to our original equation except we made a = 1 and b = 1000:

- First lets solve for our variable 'x':

- lets check our numbers to see if the total sum of sales still equals 'a' squared:

- So parc (c) of (e) checks out with the variables a = 1 and b = 1000

- [
**Part (2) of (e)**:] - We will modify our python code to include the new values of 'a' and 'b', also we had to round for our comparison statement. Python will give you an error if you don't round your total sales variable when you compare in the 'if' statement'. The code will look like this:

a = 1.0 b = 1000.0 i = 35 loop_condition = False while loop_condition == False: z = 1.0 + (1/2.0) + 1/(2.0**i) + 1.0 x = (1.0/z)*(a**2.0 + a - b) i -= 1 total_sales = (x + (x/2.0) + x + (b-a) + (x/2.0)) print 'the total sales equals: ', total_sales if round(total_sales,1) == a**2.0: print 'this is i', i loop_condition = True

- Looking at the screen results we this:

- The results show us the same results we calculated in the previous part (d), as Lynda's sales converges closer to Timmy's then the formula holds true. Know that she needs to match Timmy's sales for the month, it does not benefit to have her on the team.

- [
**Challenge**:] Can you figure out what will be optimal if a = 50 and b = 100 if only 2 salespeople are to work? How about 3? Why?

- So we have finally arrived at the part you really wish to know, to start off lets begin with our three original salesman. We are going to do the same steps:

- First lets solve for our variable 'x':

- lets check our numbers to see if the total sum of sales still equals 'a' squared:

- Our story checks out for 3 salesman, but what about only 2? Well this begs the question, who do you cut? If we put in order of sales, it would go Timmy < Mike < Joe. We could denote this notion like so:

- If we cut the smallest portion, then we eliminate Timmy. Now lest test our new equations:

- First lets solve for our variable 'x':

- lets check our numbers to see if the total sum of sales still equals 'a' squared:

- Either I got lucky and cut the right person, or there is a system. So currently we only needed Mike and Joe's sales for the conjecture to hold true. This leads me to investigate what happens when I take out the bigger sales from the equation. So lets cut Joe's sales from our current equation:

- First lets solve for our variable 'x':

- lets check our numbers to see if the total sum of sales still equals 'a' squared:

- It seems our findings are still good, did not even need the extra variable (b). There is one last person to cut, we need to see what happens if we cut Mike's sales. This will help us understand even more about this system.

- First lets solve for our variable 'x':

- lets check our numbers to see if the total sum of sales still equals 'a' squared:

- [
**Conclusion**:] So, we have cut all three salesman from the equation and the total sales still equaled 'a' squared. It is more optimal to only take two of the salesman than it is to have all three. This is not the first time I ran into this type of problem. Back in college I encountered a similar problem (different names) but the same math was used with the same variables. I believe the explanation given to me at the time was the reciprocal rule. I will leave it to you to verify such an explanation to your question. This is as far as I can currently take you. I hope there was some useful information to your question.

Reference:

- Valdez-Sanchez, Luis. "Table of Trigonometric Identities ."
*S.O.S Math*. N.p., 3 Nov. 1996. Web. 30 June 2013. <http://www.sosmath.com>. Path: http://www.sosmath.com/trig/Trig5/trig5/trig5.html. *Wikipedia, the free encyclopedia*. N.p., n.d. Web. 30 June 2013. <http://en.wikipedia.org>. Path: http://en.wikipedia.org/wiki/Pythagorean_trigonometric_identity.*Wikipedia, the free encyclopedia*. N.p., n.d. Web. 30 June 2013. <http://en.wikipedia.org>. Path: http://en.wikipedia.org/wiki/Reciprocal_rule.