Question 175440
Given A can do the work in x hours and B can do the work in y hours, you proceed by saying A can do {{{1/x}}}th of the job in 1 hour and B can do {{{1/y}}}th of the job in 1 hour, so together they can do {{{1/x+1/y}}}th of the job in 1 hour and therefore can do the whole job in {{{1/(1/x+1/y)}}} hours.  The above can be extended to as many workers as you like just by adding terms.


So for your problem:


A can do the work in {{{t + 1}}} hours, therefore can do {{{1/(t+1)}}} of the job in 1 hour.


B can do the work in {{{t + 6}}} hours, therefore can do {{{1/(t+6)}}} of the job in 1 hour.


C can do the work in {{{2t}}} hours, therefore can do {{{1/2t}}} of the job in 1 hour.


Together they can do the whole job in {{{t}}} hours, so they can do {{{1/t}}} of the job in 1 hour.


Therefore:


{{{(1/(t + 1)) + (1/(t + 6)) + (1/2t) = 1/t}}}


This looks like an algebraic horror, but really isn't that bad.  First, find the common denominator.  The factors of the common denominator are {{{(t + 1)}}}, {{{(t + 6)}}}, {{{2t}}}, and {{{t}}}, and only the {{{2t}}} and {{{t}}} have the common factor {{{t}}}, so we can eliminate that making the LCD the product:  {{{ (t + 1)(t + 6)(2t) }}}.   It will be convenient to leave the LCD in that form as you will see.


Applying the LCD we get:


{{{(((t + 6)(2t)) + ((t + 1)(2t)) + ((t + 1)(t + 6)))/((t + 1)(t + 6)(2t))=(2(t + 1)(t + 6))/((t + 1)(t + 6)(2t))}}} 


The first thing to do to simplify this mess is to multiply both sides of the equation by the denominator leaving the two numerator expressions equal:


{{{((t + 6)(2t)) + ((t + 1)(2t)) + ((t + 1)(t + 6))=2(t + 1)(t + 6)}}} 


Next, perform the indicated multiplications:


{{{(2t^2 + 12t) + (2t^2 + 2t) + (t^2 + 7t + 6) = 2t^2 + 14t + 12}}}


Remove the parentheses:


{{{2t^2 + 12t + 2t^2 + 2t + t^2 + 7t + 6 = 2t^2 + 14t + 12}}}


Collect terms on the left:


{{{5t^2 + 21t + 6 = 2t^2 + 14t + 12}}}


Add {{{-(2t^2 + 14t + 12)}}} to both sides:


{{{3t^2 + 7t - 6 = 0}}}


Leaving us with a not too unattractive little quadratic to solve.  I say not too unattractive because the thing actually factors:


{{{(3t - 2)(t + 3) = 0}}}


So:  {{{3t - 2 = 0}}} → {{{t = 2/3}}} or {{{t + 3 = 0}}} → {{{t = -3}}}


Notice that {{{t = -3}}} is an absurd answer because that would mean if they started the job now, they would be done three hours ago.  What happened is that in the process of solving the problem we squared the variable and that introduced an extraneous root.  Exclude {{{t = -3}}}, leaving us with {{{t = 2/3}}} as our solution set.


Check the answer:


A can do the work in {{{2/3 + 1}}} hours, therefore can do {{{1/(2/3 + 1)}}} of the job in 1 hour.


B can do the work in {{{2/3 + 6}}} hours, therefore can do {{{1/(2/3 + 6)}}} of the job in 1 hour.


C can do the work in {{{2(2/3)}}} hours, therefore can do {{{1/(2(2/3))}}} of the job in 1 hour.


Together they can do the whole job in {{{2/3}}} hours, so they can do {{{3/2}}} of the job in 1 hour.



{{{(1/(5/3)) + (1/(20/3)) + (1/(4/3)) = 3/2}}}



{{{3/5 + 3/20 + 3/4 = 3/2}}}


{{{12/20 + 3/20 + 15/20 = 30/20 = 3/2}}} Answer checks.


<large><b>Super Double Plus Extra Credit</large></b>


Say that you are paying each of these three guys $15.00 per hour and your customer is paying you $75 to get the truck loaded.


1.  What is your profit given the stated conditions?


2.  What action could you take to maximize your profit (aside from doing the whole job yourself)?  By how much?