The Encyclopedia of Economic Principals

Chapter 3

Production Duality, Factor Demand, and Firm Optimization

Choose output at the margin and inputs at least cost.

Four of the chapter's worked examples, made interactive: a monopolist's output with and without a capacity limit, Shephard's lemma, the shutdown rule, and least-cost inputs under a labor minimum.

Every example here is a constructed teaching example: it uses the hypothetical numbers of the chapter's worked examples, plus a few values added for comparison and labelled as such in each panel. Nothing here measures a real market, firm or household.

Demonstration 1 of 4

MR equals MC, then a capacity ceiling

Where does a monopolist set output, and what does a binding capacity limit leave on the table?

Marginal revenue falls with output and marginal cost rises, so profit peaks where they cross. If capacity stops the firm short of that point, it produces at capacity and the remaining gap MR - MC measures what one more unit of capacity is worth.

Equation, written in LaTeX: R(q)=120q-2q^2, MR(q)=120-4q, MC(q)=20+2q.

Equation, written in LaTeX: 120-4q=20+2q \Longrightarrow q^*=\frac{50}{3}\approx16.67.

Equation, written in LaTeX: MR(12)-MC(12)=72-44=28.

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q is output. Inverse demand is P(q) = a - 2q with intercept a (120 in the book) and total cost is C(q) = 200 + 20q + q^2. MR is marginal revenue and MC marginal cost.

Predict first. Does a ceiling at 16 change profit by much compared with a ceiling at 12?

Your prediction

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Figure: MR equals MC, then a capacity ceiling. Marginal revenue and marginal cost lines against output with demand intercept 120. The chosen output is 16.67, where the two lines cross.
Demand intercept: 120, Capacity ceiling: None
Constructed example: the chapter's hypothetical monopolist (intercept 120, no cap and a cap of 12); intercepts 100 and 140 and caps of 16 and 8 are added for comparison.

Calculated values

Output
16.67
Price
$86.67
Revenue
$1,444.44
Total cost
$811.11
Profit
$633.33
MR - MC at the chosen output
0.00

120 - 4q = 20 + 2q gives q = (120 - 20) / 6 = 50/3 = 16.67. Price = 120 - 2 x 50/3 = 260/3 = 86.67; revenue = 260/3 x 50/3 = 13,000/9 = 1,444.44; cost = 200 + 20 x 50/3 + (50/3)^2 = 7,300/9 = 811.11; profit = 13,000/9 - 7,300/9 = 1,900/3 = 633.33.

Worked steps

  1. Unconstrained: q = (120 - 20) / 6 = 50/3 = 16.67
  2. Capacity does not bind
  3. Price = 120 - 2 x 50/3 = 260/3 = 86.67
  4. Revenue = 260/3 x 50/3 = 13,000/9 = 1,444.44
  5. Cost = 200 + 20 x 50/3 + (50/3)^2 = 7,300/9 = 811.11
  6. Profit = 13,000/9 - 7,300/9 = 1,900/3 = 633.33

Use the idea

Use the MR - MC gap at a binding limit as the most a firm should pay for one more unit of capacity, holding demand and cost fixed.

Where the conclusion applies

Known linear demand, quadratic cost and continuous output. The fixed cost of 200 changes profit but not the chosen output.

Check your understanding: With demand intercept 140 and no cap, what output and profit?
140 - 4q = 20 + 2q gives q = 20 and price 100; profit = 2,000 - (200 + 400 + 400) = 1,000.

Chapter 3 source: section "Profit-maximization condition".

Demonstration 2 of 4

Cost function slopes are input demands

Why does the slope of the cost function with respect to the wage equal the labor the firm hires?

Shephard's lemma says the wage derivative of minimum cost is the cost-minimizing labor demand. The cost curve is concave in the wage, so its tangent lies above it: holding labor fixed overstates the cost of a wage rise.

Equation, written in LaTeX: L^c=y\sqrt{\frac{r}{w}}, K^c=y\sqrt{\frac{w}{r}}, c(w,r,y)=2y\sqrt{wr}

Equation, written in LaTeX: \frac{\partial c}{\partial w}=y\sqrt{\frac{r}{w}}=50, \frac{\partial c}{\partial r}=y\sqrt{\frac{w}{r}}=200

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Technology is y = sqrt(LK) with y = 100 units. w is the wage, r the price of machine services, L and K the least-cost inputs and c(w, r, y) the minimum cost.

Predict first. Does the tangent line over- or under-predict cost after a wage rise?

Your prediction

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Figure: Cost function slopes are input demands. The minimum cost curve against the wage with machine price 4, and its tangent at w = 16. At w = 16 cost is 1,600.00 and the tangent gives 1,600.00.
Wage: 16, Price of machine services: 4
Constructed example: the chapter's hypothetical firm (y = 100, r = 4, wage 16 then 25); wages of 9 and 36 and machine prices of 1 and 9 are added for comparison.

Calculated values

Labor L
50.00
Machine services K
200.00
Minimum cost
1,600.00
Tangent prediction from w = 16
1,600.00
Tangent minus actual
0.00

L = 100 x sqrt(4 / 16) = 50.00 and K = 100 x sqrt(16 / 4) = 200.00, so cost = 2 x 100 x sqrt(16 x 4) = 1,600.00, which equals 16 x 50.0000 + 4 x 200.0000. Holding labor at its w = 16 level predicts 1,600.00 + 50.00 x (16 - 16) = 1,600.00, exactly the cost at w = 16.

Worked steps

  1. L = 100 x sqrt(4 / 16) = 50.00
  2. K = 100 x sqrt(16 / 4) = 200.00
  3. Cost = 2 x 100 x sqrt(64) = 1,600.00
  4. Tangent = 1,600.00 + 50.00 x 0 = 1,600.00
  5. Tangent - actual = 1,600.00 - 1,600.00 = 0.00

Use the idea

Read input demands off an estimated cost function, and treat a fixed-input cost forecast as an upper bound when the firm can substitute.

Where the conclusion applies

Cost minimization with technology sqrt(LK) and price-taking in both inputs. The book's tangent comparison uses r = 4; with other r the same logic holds at that r.

Check your understanding: At w = 36 and r = 4, what labor demand and cost?
L = 100 x sqrt(4 / 36) = 33.33 and cost = 2 x 100 x sqrt(144) = 2,400.

Chapter 3 source: section "Shephard's lemma".

Demonstration 3 of 4

Operate at a loss or shut down?

When should a plant that loses money keep running?

Only avoidable costs matter for the run-or-close decision. The plant runs when revenue covers avoidable cost, that is when price is above the minimum of AVC. The $500 commitment is paid either way.

Equation, written in LaTeX: VC(q)=80+10q+q^2.

Equation, written in LaTeX: AVC(q)=\frac{80}{q}+10+q, MC(q)=10+2q.

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q is monthly output and p the market price. VC(q) is avoidable cost when the plant opens, including an $80 opening cost; $500 is paid whether or not it operates. AVC is average variable cost and MC marginal cost.

Predict first. At price 28, just above the shutdown price of 27.89, does the plant run?

Your prediction

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Figure: Operate at a loss or shut down? Average avoidable cost and marginal cost curves with a price line at 30. At q = 10 the contribution is 20, so the plant should operate.
Market price: 30, Is the $80 opening cost avoidable?: Yes
Constructed example: the chapter's hypothetical plant (prices 30 and 26, opening cost avoidable or owed); prices 24 and 28 are added for comparison.

Calculated values

Output if open
10
Revenue
$300
Avoidable cost
$280
Contribution
$20
Profit if open
-$480
Profit if closed
-$500
Decision
operate

Price 30 = 10 + 2q gives q = 10. Revenue = 30 x 10 = 300; avoidable cost = 80 + 10 x 10 + 10 x 10 = 280; contribution = 300 - 280 = 20. Open, profit is -480; closed, -500. The plant should operate.

Worked steps

  1. q from 30 = 10 + 2q: q = 10
  2. Revenue = 30 x 10 = 300
  3. Avoidable cost = 80 + 100 + 100 = 280
  4. Contribution = 300 - 280 = 20
  5. Decision: operate

Use the idea

Before closing a loss-making operation, compare revenue with the costs that closing would actually avoid this period.

Where the conclusion applies

A price taker with known costs and a one-month horizon. Restart costs, financing or expectations about future prices would need a dynamic comparison.

Check your understanding: At price 28 with an avoidable opening cost, what is the contribution?
q = 9, revenue 252, avoidable cost 80 + 90 + 81 = 251, so contribution is +1 and the plant operates.

Chapter 3 source: section "Shutdown rule".

Demonstration 4 of 4

Least-cost inputs and a labor floor

How does the least-cost mix of labor and capital respond to the wage, and what does a labor minimum cost?

At the least-cost point the isocost line is tangent to the isoquant: the ratio of marginal products equals the price ratio. A higher wage tilts the isocost and moves the tangency toward capital. A labor floor can force a corner where the tangency no longer holds.

Equation, written in LaTeX: \frac{f_L}{f_K}=\frac{K}{2L}=\frac{w}{r}.

Equation, written in LaTeX: 100=L^{1/3}(8L)^{2/3}=4L,

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Technology is y = L^(1/3) K^(2/3) with a target of 100 units. w is the wage and r = 5 the price of capital services; f_L and f_K are the marginal products. The book's equations are its w = 20 case, where K = 8L.

Predict first. Does the labor floor of 40 raise cost more when wages are high or low?

Your prediction

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Figure: Least-cost inputs and a labor floor. Isoquant for 100 units with the isocost line touching it at L = 25.00, K = 200.00.
Wage: 20, Minimum labor: None
Constructed example: the chapter's hypothetical firm (wages 20 and 45, r = 5, target 100, labor minimum 40); the wage of 5 and the numbers under the floor are added for comparison.

Calculated values

Labor L
25.00
Capital K
200.00
Cost
$1,500.00
Interior K/L
8.00
Cost without the floor
$1,500.00
Labor floor
none

K/(2L) = 20/5 gives K = 8.00L, and 100 = L^(1/3) (8.00L)^(2/3) gives L = 100 / 8.00^(2/3) = 25.00, K = 200.00. Cost = 20 x 25.0000 + 5 x 200.0000 = 1,500.00.

Worked steps

  1. K = (2 x 20 / 5) L = 8.00L
  2. L = 100 / 8.00^(2/3) = 25.00
  3. K = 8.00 x 25.0000 = 200.00
  4. Cost = 20 x 25.0000 + 5 x 200.0000 = 1,500.00

Use the idea

Estimate what a staffing rule costs by comparing least cost with and without it at the current input prices.

Where the conclusion applies

Constant returns, known prices and a fixed output target. Exact values: at w = 45, L = 14.5597 and K = 262.07 to two decimals.

Check your understanding: At w = 20 with the 40 labor floor, what capital and cost?
K = (100 / 40^(1/3))^(3/2) = 158.11 and cost = 20 x 40 + 5 x 158.11 = 1,590.57.

Chapter 3 source: section "Cost Minimization and Constrained Optimization".