Question 415235
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Let *[tex \Large x] represent the number of items produced in one day.  Then *[tex \Large 8x] dollars is the variable cost based on the number of items produced and *[tex \Large 8x\ +\ 1500] must be the total cost.   The total revenue is the price times the number of items sold.  So if you let *[tex \Large y] be the number of items sold, then *[tex \Large 12y] dollars is the revenue.


The difficulty with the problem as stated is that there is no data or information given that allows us to relate the number of items produced with the number of items sold.


The breakeven point is where the total revenue is equal to the total cost:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8x\ +\ 1500\ =\ 12y].


Which is a single linear equation in two variables, having an infinite number of solutions.  This means there is no definative answer to the question.


If you assume, though you really have no right to do so given the wording of the problem, that the number sold is equal to the number produced, then you have two equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ y]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8x\ +\ 1500\ =\ 12y]


And you can solve the system by first substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8x\ +\ 1500\ =\ 12x]


and then solving by ordinary means.


However, consider the case where they sold 100 items less than they produced.  Then you would have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ 100\ =\ y]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8x\ +\ 1500\ =\ 12y]


And you can solve the system by first substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8x\ +\ 1500\ =\ 12(x\ -\ 100)]


and the solution in this case is very different.


You only get to ask one question per post.  Re-post your second question.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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