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Principles of Microeconomics · Opportunity cost and production possibilities

Use a production possibilities frontier to identify feasible output, calculate the cost of a trade-off, and explain how more resources change the frontier.

Before you start

You should be able to read ordered pairs, solve a linear inequality, and calculate slope as vertical change divided by horizontal change. Keep the units in every calculation.

A model of scarce time

A workshop has 24 labor hours per day. A stool takes 2 hours and a shelf takes 3. Let x be stools per day and y be shelves per day. With unchanged methods, nonnegative output and divisible batches, a bundle is feasible exactly when 2x + 3y ≤ 24. The inequality is necessary because time cannot exceed the budget. It is sufficient because the two tasks use time independently and any unused time may be left idle.

The production possibilities frontier (PPF) is the upper boundary of feasible output, here y = 8 − (2/3)x for 0 ≤ x ≤ 12. Each point on this boundary is productively efficient because raising either output requires lowering the other. An interior point is feasible with unused time in this model. An exterior point requires more than 24 hours. Choosing which efficient bundle is preferable requires information about preferences or objectives.

Workshop production possibilities The line connects zero stools and eight shelves to twelve stools and zero shelves. A at six stools and four shelves is on the frontier. B at six stools and two shelves is inside. C at six stools and six shelves is outside. ABC 8012 Shelves (y) Stools (x)
The shaded triangle is feasible, including its edges. A = (6, 4) is on the frontier, B = (6, 2) leaves time unused, and C = (6, 6) exceeds the time budget.

Opportunity cost and slope

Opportunity cost is the value of the best alternative forgone. On this frontier, the extra time for one stool could instead make 2/3 of a shelf, so the opportunity cost is 2/3 shelf per stool. The slope is negative because the outputs move in opposite directions. The cost is the positive amount forgone. With differing worker skills, successive reallocations can have different costs, so a frontier need not be straight.

Worked examples

  1. Classify B = (6, 2). It takes 2×6 + 3×2 = 18 hours, leaving six hours. With six stools fixed, those hours can produce two more shelves, reaching A = (6, 4).
  2. Move from A = (6, 4) to (9, 2). Three extra stools take six hours, so two shelves must be given up. Divide 2 by 3 for the cost per stool. In the reverse direction, the cost is 3/2 stools per extra shelf.
  3. Reduce shelf time from 3 hours to 2 while keeping 24 labor hours and stool time fixed. The shelf intercept rises from 8 to 12 and the stool intercept stays 12. The new frontier y = 12 − x pivots around (12, 0), and each stool now costs one shelf.

Common mistakes

  • Keep the units straight. Shelf per stool and stool per shelf are reciprocal rates.
  • Distinguish moving to the frontier by using idle time from shifting the frontier by changing resources or methods.
  • If only whole products are possible, round the maximum feasible output down and report leftover time. The line is then a continuous benchmark.

Practice

  1. Classify (3, 6) and (9, 3). Show the hours used.
  2. Start at (3, 6). On the original frontier, how many shelves remain if the shop makes 7 stools? What is the total cost in shelves?
  3. Raise available hours to 30. Derive the new frontier, identify both intercepts, and explain why slope stays the same.
Worked practice solutions
  1. (3, 6) uses 6 + 18 = 24 hours, so it is on the frontier. (9, 3) uses 18 + 9 = 27, so it is outside.
  2. Seven stools take 14 hours. The remaining 10 hours make 10/3 shelves. The sacrifice is 6 − 10/3 = 8/3 shelves, the same as 4 extra stools times 2/3 shelf per stool.
  3. Solve 2x + 3y = 30 for y: y = 10 − (2/3)x, 0 ≤ x ≤ 15. The intercepts are (0, 10) and (15, 0). The ratio of hours per product is unchanged, so the slope is still −2/3.

Mini-exam · 10 points

A second shop has 36 hours. A stool takes 3 hours and a shelf takes 4.

  1. (4 points) Write the feasible inequality with domain conditions and give both intercepts.
  2. (3 points) Classify (4, 5) and state the maximum whole shelves with four stools.
  3. (3 points) On the frontier, how many shelves does moving from four to eight stools cost? Give the cost per extra stool with units.
Answer key and scoring
  1. One point for 3x + 4y ≤ 36, one for x, y ≥ 0, and one each for (12, 0) and (0, 9).
  2. One point each for 32 hours used, the interior classification, and (36 − 12)/4 = 6 shelves.
  3. One point for identifying six shelves initially and three afterward, one for the loss of three shelves, and one for 3/4 shelf per extra stool.

Free reading

OpenStax · Principles of Economics 3e, §2.2

Virtual lab · Reallocate workshop time

Use a fixed time budget to compare feasible, efficient, and infeasible output bundles, then test a resource increase.

Free tool built into this page. Use it in your browser without an account or download.

Procedure

Use the three inputs below. Set the budget to 24 hours and hold stools at 6. Predict time used and time remaining for 2, 4, and 6 shelves, then run each case and record the classification. Keep the last bundle and raise the budget to 30. Why does its classification change?

Expected results and answer check

At 24 hours, the bundles use 18, 24, and 30 hours, leaving +6, 0, and −6. They are inside, on, and outside the frontier. A 30-hour budget puts the last bundle on the new frontier because 2×6 + 3×6 = 30. The slope stays −2/3 because unit labor requirements are unchanged.

Inputs are whole counts but classification is relative to the continuous frontier. A small amount of idle time may be insufficient to make a whole extra item. The activity uses synthetic labor requirements and does not measure real productivity, setup time, or bottlenecks.

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