Answer:
a) optimum number of days for the job to last = 24 days
b) Total income for the optimum number of days = $2434.83
c) Total expenditures for the optimum number of days = $341.77
d) Maximum profit for the job = $2093.06
Step-by-step explanation:
a) Expenditure rate ,[tex]E'(x) = e^{0.17x}[/tex]
Income rate [tex]I'(x) = 116.2-e^{0.17x}[/tex]
The job lasts for many days where the income is more than expenditure and job ends when both are equal.
The optimum number of days is the value of x where I'(x) = E'(x)
[tex]116.2-e^{0.17x} =e^{0.17x}\\116.2=2e^{0.17x}\\58.1=e^{0.17x}\\ {0.17x} = ln(58.1)[/tex]
[tex]x = ln(58.1)/0.17\\x = 23.9[/tex]
The optimum number of days for the job to last is 24 days
b) [tex]I'(x) = 116.2-e^{0.17x}[/tex]
Integrating both sides in the equation above:
[tex]I(x) = 116.2x-(e^{0.17x}/0.17)[/tex]
Substituting [tex]x = ln(58.1)/0.17[/tex] into the equation above , The income for the optimum number of days becomes:
[tex]I(x) = 116.2\times (ln(58.1)/0.17)-(e^{0.17(ln(58.1)/0.17)}/0.17)I(x) = (116.2\times (ln(58.1)/0.17))-(58.1)/0.17)I(x)=2776.60 -341.76I(x) =2434.83[/tex]
c)
[tex]E'(x) = e^{0.17x}[/tex]
Integrating the equation above:
[tex]E(x) = (1/0.17)e^{0.17x}[/tex]
Substituting [tex]x = ln(58.1)/0.17[/tex] into the above equation, expenditure for the optimum number of days is;
[tex]E(x) =1/(0.17)e^{0.17(ln(58.1)/0.17)} \\E(x) =58.1/0.17\\E(x)=341.77[/tex]
d) Maximum profit is 2434.83- 341.77=2093.06