Question

The owner of a photocopy store charges 9 cents per copy for the first 100 copies and 6 cents per copy for each copy exceeding 100. In​ addition, there is a setup fee of ​$1.60 for each photocopying job.

​(a) Determine​ R(x), the revenue from doing one photocopying job consisting of x copies.
​(b) If it costs the store owner 4 cents per​ copy, what is the profit from doing one photocopying job consisting of x​ copies?

Answer 1

a) The revenue function, in $, is given by

`R(x) = \left \{ {1.60 + 0.09*x \,\, \, {x \leq 100} \atop 10.60 + 0.06*x \,\,\, {x >100}} \right`

b) The profit function is $ is given by

`p(x) = \left \{ {1.6 + 0.05*x \, \, {x \leq 100} \atop 10.6+0.02 *x \, \, {x>100}} \right`

Explanation:

a) For a value of x less than or equal to 100, we have that the revenue function (in $) is

`R(x) = 1.60 + 0.09 * x` .

For x = 100 therefore, we have that

`R(100) = 1.60 + 0.09*100 = 10.60`

And for x = 100+k, for k positive, we have that the first 100 copies will cost as usual, but the remaining only will cost $0.06, thus

`R(x) = R(100) + 0.06*x = 10.60 + 0.06 *x`

With this information we can conclude that

`R(x) = \left \{ {1.60 + 0.09*x \,\, \, {x \leq 100} \atop 10.60 + 0.06*x \,\,\, {x >100}} \right.`

(Note that x should satisfy x≥0, otherwise the function woudlnt make sense)

b) The profit function p(x) can be obtained from R(x) by substracting the cost of making x copies, which is
`0.04*x` . This way, p(x) = R(x) -0.04x is given as follows

`p(x) = \left \{ {1.6 + 0.05*x \, \, {x \leq 100} \atop 10.6+0.02 *x \, \, {x>100}} \right.`