Costs of Production

1. Costs of Production

2. In the previous section, we looked at production. In this section, we look at the cost of production & determining the optimal level of output. We’ll start with an interesting cost example, & then focus on determining the optimal level of output in general.

3. Optimal Level of Law Enforcement & Crime Prevention. Consider the total cost of crime (CT), including both the cost of the criminal act itself (CA) & the cost of law enforcement & crime prevention (CP). We’d like to know the optimal level of law enforcement & crime prevention (L) that will minimize the total cost CT = CP + CA . Remember from calculus that to minimize a function, we take the first derivative and set it equal to zero. So we will have dCT/dL = dCP/dL + dCA/dL = 0 or dCP/dL = – dCA/dL . This means that the optimal enforcement level is where the cost of preventing an additional crime is equal to the cost of an additional criminal act. To finish solving for the optimal level (L), we’d need to know the specific form of the cost functions involved.

4. Isocost Curve • The set of combinations of inputs that cost the same amount

5. Equation of an isocost • Suppose you have 2 inputs, capital K & labor L. • The price of a unit of capital is PK. • The price of a unit of labor is PL. • Let a particular outlay amount be R. • Then all combinations of K & L such that PLL+ PKK = R lie on the isocost curve associated with that outlay. • If we rewrite the equation as K = R/PK – (PL/PK)L ,we see that the slope of the isocost is – (PL/PK) & the vertical intercept is R/PK .

6. Graph of an isocost For example, suppose you’re interested in the outlay amount $10,000. Suppose also that Labor cost $10 per unit & capital cost $100 per unit. Then the slope of the isocost is – PL/PK = – 10/100 = – 0.1 . The vertical intercept would be 10,000/100 = 100 & the horizontal intercept is 10,000/10 = 1,000. K R/PK= 100 slope = – PL/PK = – 0.1 R/PL = 1000 L

7. Maximizing output for a given cost level At points A & B, we’re spending the outlay associated with this isocost, but we’re not producing as much as we can. We’re only making Q1 units of output. We can’t produce Q3 or Q4 with this outlay. Those output levels would cost more. At point E, we’re producing the most for the money, where the isocost is tangent to an isoquant. isoquants K A isocost E Q4 Q3 Q2 B Q1 L

8. At the tangency of the isocost & isoquant, the slopes of those curves are equal. We found previously that the slope of the isoquant is – MPL/MPK , & the slope of the isocost is – PL/PK . So at the tangency, – MPL/MPK = – PL/PK or, multiplying by -1, MPL/MPK = PL/PK . This expression is equivalent to MPL/PL = MPK/PK .

9. MPL/PL = MPK/PK This condition means that to get the most output for your money, you should employ inputs such that the marginal product per dollar is equal for all inputs. (Notice the similarity to the utility maximization condition that the marginal utility per dollar is equal for all goods.)

10. Minimizing cost for a given output level At points A & B, we’re producing the desired quantity, but we’re not using the cheapest combination of inputs, so we’re spending more than necessary. We can’t produce the desired output level at cost level C1. We need more money. At point E, we’re producing the desired output at the lowest cost, where the isoquant is tangent to an isocost. K A E isoquant Q1 B L isocosts C1 C2 C3

11. So whether we’re maximizing output for a given cost level, or minimizing cost for a given output level, the condition is the same: • MPL/PL = MPK/PK • The marginal product per dollar is equal for all inputs.

12. Short Run Costs of Production

13. Total Fixed Cost (TFC) Total fixed cost is the cost associated with the fixed input.

14. $ TFC Quantity Since TFC is constant, its graph is a horizontal line.

15. Average Fixed Cost (AFC) AFC = TFC/Q AFC is the fixed cost per unit of output.

16. The AFC curve slopes downward & gets closer & closer to the horizontal axis. $ AFC Quantity

17. Total Variable Cost (TVC) Total variable cost is the cost associated with the variable input.

18. TVC $ Quantity The TVC curve is upward sloping. It is often drawn like a flipped over S, first getting flatter & flatter, & then steeper & steeper. This shape reflects the increasing & then decreasing marginal returns we discussed in the section on production.

19. Average Variable Cost (AVC) AVC = TVC/Q AVC is the variable cost per unit of output.

20. We can determine the shape of the AVC curve based on the shape of the average product curve (AP). Suppose X is the amount of variable input & PX is its price. Then, AVC = TVC/Q = (PXX)/Q = PX(X/Q) = PX [1/(Q/X)] = PX [1/AP]. So since AP had an inverted U-shape, AVC must have a U-shape.

21. Average Variable Cost $ AVC Quantity

22. TC $ TFC Total Cost TC = TFC + TVC The TC curve looks like the TVC curve, but it is shifted up, by the amount of TFC. Quantity

23. Average Total Cost Like AVC, ATC is U-shaped, but it reaches its minimum after AVC reaches its minimum. This is because ATC = AVC +AFC & AFC continues to fall & pulls down ATC. ATC $ AVC Quantity

24. Marginal Cost (MC) MC is the additional cost associated with an additional unit of output. MC = ΔTC/ ΔQ Alternatively, MC = dTC/dQ . MC is the first derivative of the TC curve or the slope of the TC curve.

25. We can determine the shape of the MC curve based on the shape of the marginal product curve (MP). Suppose the firm takes the prices of inputs as given. Then,MC = TC/Q = PX X/ Q = PX [1/(Q/X)] = PX [1/MP]. So since MP had an inverted U-shape, MC must have a U-shape.

26. MC $ Quantity While MC is U-shaped, it is often drawn so it extends up higher on the right side.

27. MC ATC $ AVC Quantity Important Graphing Note: The MC must intersect the ATC at its minimum & the AVC curve at its minimum.

28. We have a similar graphical interpretation of ATC to the one we had for AP. Since ATC = TC/Q, the ATC of a particular value of Q1 can be interpreted as the slope of the line from the origin to the corresponding point on the curve. TC TC  TC1  0  Q1 → Q

29. We also have similar graphical interpretation of MC to the ones we had for MP. The continuous MC is the slope of the total cost curve at a particular point. The discrete MC is the slope of the line segment connecting 2 points on the total cost curve. TC TC Q

30. TC TR TC Q Breaking Even Recall that TR = PQ. If the price of output is fixed for the firm (as for a perfectly competitive firm), then TR is a straight line with slope P. When the TR curve is above the TC curve, the firm will have positive economic profits. When the TC curve is above the TR curve, the firm will have economic losses. The firm will break even (have zero economic profits) where TR=TC.

31. TC TR TC Profit-maximizing output level Q Maximizing Profit The firm will have the maximum profits where the vertical distance between TR & TC is the largest (& TR is above TC). This is also where MR = MC (which you should recall from Micro Principles is the profit maximizing condition). That means that the slope of the TR line equals the slope of the TC curve. So the TR line will be parallel to a tangent to the TC line at the point where profits are maximized.

32. TC TR TC Profit-minimizing output level Q Minimum Profit Notice that the TR line is also parallel to a tangent to the TC line here. TR – TC reaches a minimum here, not a maximum.

33. We’re going to digress a little to review from Calculus how to use first and second derivatives to determine minima and maxima. y Consider a function y = f(x) as shown. Notice that it has a minimum value at x1. Notice also that the slope of the function (which is the same as the slope of the line tangent to the curve at that point) is zero. That is, f (x1) = 0. x1 x

34. y f  (x) > 0 f  (x) < 0 Just to the left of x1, the curve slopes downward; it has a negative slope. To the right of x1, the curve slopes upward; it has a positive slope. x1 x

35. y f  (x) < 0 f  (x) > 0 So as we move from left to right in the vicinity of x1, the slope is going from negative to zero to positive. It is increasing. Recall that if a function is increasing, its derivative is positive. In this case, the function itself is the slope or first derivative. So its derivative is the second derivative. Then, because the first derivative is increasing, the second derivative must be positive: f (x1) > 0. To put all this together: At a minimum x1 ,the first derivative f (x1) = 0 and the second derivative f (x1) > 0 . f  (x1) = 0 x1 x

36. Consider instead this function. At x1, we have a maximum. The derivative f (x1) = 0. Here, as we move from left to right in the vicinity of x1, the slope is going from positive to zero to negative. The slope is decreasing. If a function is decreasing, its derivative is negative. Again here, the function is the slope or first derivative. So its derivative is the second derivative. Then, because the first derivative is decreasing, the second derivative must be negative: f (x1) < 0. To put all this together: At a maximum x1 ,the first derivative f (x1) = 0 and the second derivative f (x1) < 0 . y f  (x1) = 0 f  (x) > 0 f  (x) < 0 x1 x

37. To summarize our conclusions on first and second derivatives and maxima and minima: At a minimum x1,the first derivative f (x1) = 0 and the second derivative f (x1) > 0 . At a maximum x1,the first derivative f (x1) = 0 and the second derivative f (x1) < 0 .

38. Memory Aid If the second derivative is positive, we have two happy twinkly eyes and a smiling mouth which has a minimum. If the second derivative is negative, we have two sad eyes and a sad mouth which has a maximum. Let’s return to maximizing profit and see how we use our Calculus in this context.

39. Example: Suppose the price of a product is $10. The cost of production is TC = Q3 – 21Q2 + 49Q+100.What is the profit maximizing output level? We need to determine the profit function , take its 1st derivative, set that equal to zero, & solve for Q. = TR –TC = PQ – TC = 10Q – (Q3 – 21Q2 + 49Q+100) = 10Q – Q3 + 21Q2 – 49Q – 100 = – Q3 + 21Q2 – 39Q – 100 d/dQ = – 3Q2 + 42Q – 39.

40. Setting the 1st derivative equal to zero we have – 3Q2 + 42Q – 39 = 0 This equation can be solved either by the quadratic formula or factoring. 1. Quadratic formula:

41. – 3Q2 + 42Q – 39 = 0 2. Factoring: – 3 (Q2 – 14Q +13) = 0 – 3 (Q – 1)(Q – 13) = 0 So either Q -1 = 0 or Q -13 = 0 , & Q = 1 or Q = 13, which is what we found by the quadratic formula. Are these both relative maxima, minima, or one of each? We need to look at the 2nd derivative of our profit function.

42. We had = – Q3 + 21Q2 – 39Q – 100 d/dQ = – 3Q2 + 42Q – 39 The 2nd derivative is – 6Q + 42 To determine whether profit is maximized or minimized at our values of 1 and 13, we need to know if the second derivative is positive or negative at each of those values. When Q = 1, – 6Q + 42 = 36 > 0 which means that  is a minimum when Q =1 . When Q = 13, – 6Q + 42 = – 36 < 0 which means that  is a maximum when Q =13 .

43. What are our maximum & minimum profit values? • = – Q3 + 21Q2 – 39Q – 100 Our maximum , which is when Q = 13, is:  = – (13)3 + 21(13)2 – 39(13) – 100 = 745 Our minimum , which is when Q = 1, is:  = – (1)3 + 21(1)2 – 39(1) – 100 = – 119

44. Graph of the Profit Function  = – Q3 + 21Q2 – 39Q – 100

45. Long Run Costs of Production

46. The Long Run ATC Curve(or the planning curve) • shows the least per unit cost at which any output can be produced after the firm has had time to make all appropriate adjustments in its plant size.

47. Cost SRATC1 At a relatively low output level, in the short run, the firm might have SRATC1 curve as its short run average cost curve. Quantity of output

48. Cost SRATC2 At a slightly higher output level, in the short run, the firm might have SRATC2 curve as its short run average cost curve. Quantity of output

49. Cost SRATC3 At a still higher output level, in the short run, the firm might have SRATC3 curve as its short run average cost curve. Quantity of output

50. Cost LRATC SRATC1 SRATC5 SRATC2 SRATC4 SRATC3 In the long run, the firm can pick any appropriate plant size. At each output level, the firm picks the plant that has the SRATC curve with the lowest value. Quantity of output