Respuesta :
Answer:
a) The revenue function, in $, is given by
[tex] R(x) = \left \{ {1.60 + 0.09*x \,\, \, {x \leq 100} \atop 10.60 + 0.06*x \,\,\, {x >100}} \right[/tex]
b) The profit function is $ is given by
[tex] p(x) = \left \{ {1.6 + 0.05*x \, \, {x \leq 100} \atop 10.6+0.02 *x \, \, {x>100}} \right [/tex]
Step-by-step explanation:
a) For a value of x less than or equal to 100, we have that the revenue function (in $) is
[tex] R(x) = 1.60 + 0.09 * x [/tex] .
For x = 100 therefore, we have that
[tex] R(100) = 1.60 + 0.09*100 = 10.60 [/tex]
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
[tex] R(x) = R(100) + 0.06*x = 10.60 + 0.06 *x [/tex]
With this information we can conclude that
[tex]R(x) = \left \{ {1.60 + 0.09*x \,\, \, {x \leq 100} \atop 10.60 + 0.06*x \,\,\, {x >100}} \right.[/tex]
(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 [tex] 0.04*x [/tex] . This way, p(x) = R(x) -0.04x is given as follows
[tex]p(x) = \left \{ {1.6 + 0.05*x \, \, {x \leq 100} \atop 10.6+0.02 *x \, \, {x>100}} \right.[/tex]
