Their first employee's output increases to $5000 until he hits his maximum efficiency when he reaches 50 customers per day when it decreases at the same rate it grew before 50. When they hire a second employee their maximum efficiency will be at a total of 100 customers, afterwhich each customer reduces the efficiency of both employees again. The second employee's efficiency curve is the same as the first employee. The cost of each employee is $100 for the day. Given n number of employees, and c number of customers, find the equation which will give the net profit for the firm for any number of employees and customers. Secondly, calculate its derivative for 6 employees and any number of customers.
What economics principle does the derivative represent?
Solution:
The first employee will have an output of $5000 at 50 customers, $0 for 0 customers, and reach 0 again at 100 customers. The intercept will thus be located at (50, $5000), and the x intercepts will be located at (0, $0) and (100, $0).
Using the quadratic equation we know that 100=(-b+sqrt(b^2-4ac))/(2a) and 0=(-b+sqrt(b^2-4ac))/(2a). We also know c will be zero (since it intercepts the y axis at 0) meaning the radicand is 0 in this quadratic equation making the equation 100=(b+sqrt(b^2))/(2a) which simplifies to 100=-b/a meaning that the quadratic equation simplifies to b=-100a.
Our other piece of information tells us the graph intercepts (50, 5000) meaning 5000=a(50)^2+b(50) which simplifies to a=(-1/50)b+2.
Combining these terms means that b=-100*((-1/50)b+2) which means b=200 and a = 6666 2/3.
R(x) = -2c2 + 200c
where c is the number of customers and R is output for a single employee.
This however is not good enough since we need it to work for all numbers of employees, so we need to multiply the number of customers by the number of employees hired which will give us their efficiency curve.
R(x) = (-2(c)2 + 200c)n
where n is the number of employees.
We then need to calculate the cost of hiring the employees, which we will call function w:
w(n) = $100n
The final question is to calculate net profit, meaning that we just simply subtract wages from revenue giving us the following:
P(n) = (-2c2 + 200c)n - 100n
We can then do the second part of the question which is to find the derivative for 6 employees:
P(6) = (-2c2 + 200c)n - 100n
P(6) = (-2c2 + 200c)6 - 100(6)
P(6) = -12c2 + 1200c - 600
Using the quadratic equation we know that 100=(-b+sqrt(b^2-4ac))/(2a) and 0=(-b+sqrt(b^2-4ac))/(2a). We also know c will be zero (since it intercepts the y axis at 0) meaning the radicand is 0 in this quadratic equation making the equation 100=(b+sqrt(b^2))/(2a) which simplifies to 100=-b/a meaning that the quadratic equation simplifies to b=-100a.
Our other piece of information tells us the graph intercepts (50, 5000) meaning 5000=a(50)^2+b(50) which simplifies to a=(-1/50)b+2.
Combining these terms means that b=-100*((-1/50)b+2) which means b=200 and a = 6666 2/3.
R(x) = -2c2 + 200c
where c is the number of customers and R is output for a single employee.
This however is not good enough since we need it to work for all numbers of employees, so we need to multiply the number of customers by the number of employees hired which will give us their efficiency curve.
R(x) = (-2(c)2 + 200c)n
where n is the number of employees.
We then need to calculate the cost of hiring the employees, which we will call function w:
w(n) = $100n
The final question is to calculate net profit, meaning that we just simply subtract wages from revenue giving us the following:
P(n) = (-2c2 + 200c)n - 100n
We can then do the second part of the question which is to find the derivative for 6 employees:
P(6) = (-2c2 + 200c)n - 100n
P(6) = (-2c2 + 200c)6 - 100(6)
P(6) = -12c2 + 1200c - 600
P'(6) = -24c + 1200
This demonstrates an important point that when it comes to hiring more people assuming everything else stays constant there is a diminishing marginal utility to hiring more people, an important economics lesson.
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