A construction company has an expenditure rate of E'(x)=e^(0.17 x) dollars per day on a particular paving job and an income rate of I'(x)=116.2 - e^(.17x) dollars per day on the same​ job, where x is the number of days from the start of the job. The​ company's profit on that job will equal total income less total expenditures. Profit will be maximized if the job ends at the optimum​ time, which is the point where the two curves meet.
(a) Find the optimum number of days for the job to last.
​(b) Find the total income for the optimum number of days.
​(c) Find the total expenditures for the optimum number of days.
​(d) Find the maximum profit for the job.

Respuesta :

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

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