Question

Santa Claus is assigning elves to work an eight-hour shift making toy trucks. Apprentice earn five candy canes per hour but can only make four trucks an hour. Senior elves can make six trucks an hour and are paid eight candy canes per hour. There's only room for nine elves in the truck shop. Due to a candy-makers' strike, Santa Claus can only pay a total of 480 candy canes for the whole 8-hour shift.
(a) How many senior elves and how many apprentice elves should work this shift to maximize the number of trucks that get made?
(b) How many trucks will be made?
(c) Just before the shift begins, the apprentice elves demand a wage increase; they insist on being paid seven candy canes an hour. In order to give the apprentice elves the raise, Santa cuts back the amount of elves he could have in the truck shop to 8 elves.
Now how many senior elves and how many apprentice elves should Santa assign to this shift?

Answer 1

Answer:

  (a) 5 senior, 4 apprentice

  (b) 368 per shift

  (c) 7.5 senior, 0 apprentice

Step-by-step explanation:

The problem can be described by two inequalities. On describes the limit on the number of elves in the shop; the other describes the limit on the total payroll. Let x and y represent the number of apprentice and senior elves, respectively. Then the inequalities for the first scenario are ...

  x + y ≤ 9 . . . . . . . . . . . . total number of elves in the shop

  5x +8y ≤ 480/8 . . . . . . candy canes per hour paid to elves

These two inequalities are graphed in the first attachment. They describe a solution space with vertices at ...

  (x, y) = (0, 7.5), (4, 5), (9, 0)

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(a) Santa wants to  maximize the output of trucks, so wants to maximize the function t = 4x +6y.

At the vertices of the solution space, the values of this function are ...

  t(0, 7.5) = 45

  t(4, 5) = 46

  t(9, 0) = 36

Output of trucks is maximized by a workforce of 4 apprentice elves and 5 senior elves.

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(b) The above calculations show 46 trucks per hour can be made, so ...

  46×8 = 368 . . . trucks in an 8-hour shift

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(c) The new demands change the inequalities to ...

  x + y ≤ 8 . . . . . . number of workers

  7x +8y ≤ 60 . . . total wages (per hour)

The vertices of the feasible region for these condtions are ...

  (x, y) = (0, 7.5), (4, 4), (8, 0)

From above, we know the truck output will be maximized at the vertex (x, y) = (0, 7.5). However, we know we cannot have 7.5 senior elves working in the shop. We can have 7 or 8 elves working.

If the workforce must remain constant, truck output is maximized by a workforce of 7 senior elves.

If the workforce can vary through the shift, truck output is maximized by adding one more senior elf in the shop for half a shift.

Santa should assign 7 senior elves for the entire shift, and 8 senior elves (one more) for half a shift.

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Comment on apprentice elf wages

At 5 candy canes for 4 trucks, apprentice elves produced trucks for a cost of 1.25 candy canes per truck. At 8 candy canes for 6 trucks, senior elves produced trucks for a cost of about 1.33 candy canes per truck. The reason for employing senior elves in the first scenario is that their productivity is 1.5 times that of apprentice elves while their cost per truck is about 1.07 times that of apprentice elves.

After the apprentice elves wages were increased, their cost per truck is 1.75 candy canes per truck, but their productivity hasn't changed. They have essentially priced themselves out of a job, because they are not competitive with senior elves.