CHAPTER • The Cost of Production 249 When the effluent fee is imposed, the cost of wastewater increases from $10 per gallon to $20: For every gallon of wastewater (which costs $10), the firm has to pay the government an additional $10 The effluent fee therefore increases the cost of wastewater relative to capital To produce the same output at the lowest possible cost, the manager must choose the isocost line with a slope of −$20/$40 = −0.5 that is tangent to the isoquant In Figure 7.5, DE is the appropriate isocost line, and B gives the appropriate combination of capital and wastewater The move from A to B shows that with an effluent fee the use of an alternative production technology that emphasizes the greater use of capital (3500 machine-hours) and less production of wastewater (5000 gallons) is cheaper than the original process, which did not emphasize recycling Note that the total cost of production has increased to $240,000: $140,000 for capital, $50,000 for wastewater, and $50,000 for the effluent fee We can learn two lessons from this decision First, the more easily factors can be substituted in the production process—that is, the more easily the firm can deal with its taconite particles without using the river for waste treatment—the more effective the fee will be in reducing effluent Second, the greater the degree of substitution, the less the firm will have to pay In our example, the fee would have been $100,000 had the firm not changed its inputs By moving production from A to B, however, the steel company pays only a $50,000 fee Cost Minimization with Varying Output Levels In the previous section we saw how a cost-minimizing firm selects a combination of inputs to produce a given level of output Now we extend this analysis to see how the firm’s costs depend on its output level To this, we determine the firm’s cost-minimizing input quantities for each output level and then calculate the resulting cost The cost-minimization exercise yields the result illustrated by Figure 7.6 We have assumed that the firm can hire labor L at w = $10/hour and rent a unit of capital K for r = $20/hour Given these input costs, we have drawn three of the firm’s isocost lines Each isocost line is given by the following equation: C = ($10/hour)(L) + ($20/hour)(K) In Figure 7.6 (a), the lowest (unlabeled) line represents a cost of $1000, the middle line $2000, and the highest line $3000 You can see that each of the points A, B, and C in Figure 7.6 (a) is a point of tangency between an isocost curve and an isoquant Point B, for example, shows us that the lowest-cost way to produce 200 units of output is to use 100 units of labor and 50 units of capital; this combination lies on the $2000 isocost line Similarly, the lowest-cost way to produce 100 units of output (the lowest unlabeled isoquant) is $1000 (at point A, L = 50, K = 25); the least-cost means of getting 300 units of output is $3000 (at point C, L = 150, K = 75) The curve passing through the points of tangency between the firm’s isocost lines and its isoquants is its expansion path The expansion path describes the combinations of labor and capital that the firm will choose to minimize costs at each output level As long as the use of both labor and capital increases with output, the curve will be upward sloping In this particular case we can easily calculate the slope of the line As output increases from 100 to 200 units, capital increases from 25 to 50 units, while labor increases from 50 to 100 units For each level of output, the firm uses half as much capital as labor Therefore, the expansion path is a straight line with a slope equal to ⌬K/⌬L = (50 - 25)/(100 - 50) = • expansion path Curve passing through points of tangency between a firm’s isocost lines and its isoquants