(4,10) (8,14) Sometimes we can’t observe the production function in green. But we can observe how production changes between two discrete points. Slope.

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Presentation transcript:

(4,10) (8,14) Sometimes we can’t observe the production function in green. But we can observe how production changes between two discrete points. Slope of line between the two points is 1 In these cases, we measure marginal products by the rise/run between these two points. lbs nitrogen YieldYield 5.3. Marginal Product

Produce Wheat? Change in N (lbs / acre) Change in Wheat yield (bushels / acre)MP NO----- Yes0 →1023→ Yes10→2025→ Yes20→3030.5→ Yes30→4033→ Yes40→5035→ Yes50→6036→ MP = rise/run Going from 0 to 10 lbs N: MP = (25-23)/(10-0) = 2/10 = 0.2 MP = rise/run Going from 0 to 10 lbs N: MP = (25-23)/(10-0) = 2/10 = 0.2

5.3. Marginal Product Produce Wheat? Change in N (lbs / acre) Change in Wheat yield (bushels / acre)MP NO----- Yes0 →1023→ Yes10→2025→ Yes20→3030.5→ Yes30→4033→ Yes40→5035→ Yes50→6036→ Stage 1 Stage 2 There is no Stage 3 1. Stage 1 begins at 0→10 lbs of nitrogen per acre and ends at 10→20 lbs. 2. Stage 2 begins at 10→20 lbs of nitrogen per acre and ends at 50→60 lbs. 3. Stage 3 begins at __________ lbs of nitrogen per acre and ends at ___________ lbs. There is no Stage 3 Note that the end of Stage 1 and beginning of Stage 2 can overlap.

5.3. Marginal Product Produce Wheat? Change in N (lbs / acre) Change in Wheat yield (bushels / acre)MP NO----- Yes0 →1023→ Yes10→2025→ Yes20→3030.5→ Yes30→4040→ Yes40→5035→ Yes50→6033→ Stage 1 Stage 3 There is no Stage 2 4. Stage 1 begins at 0→10 lbs of nitrogen per acre and ends at 20→30 lbs. 5. Stage 2 begins at _____lbs of nitrogen per acre and ends at _____ lbs. There is no Stage Stage 3 begins 30→40 lbs of nitrogen per acre and ends at 50→60 lbs. Note that Stage 3 never overlaps with any other stage. Observe that this violates Stage 3, where MP is negative and should be falling (not rising) always.

5.3. Marginal Product Produce Wheat? Change in N (lbs / acre) Change in Wheat yield (bushels / acre)MP NO----- Yes0 →1023→ Yes10→2025→ Yes20→3030.5→ Yes30→4040→ Yes40→5045→ Yes50→6047→ Stage 1 Stage 3 7. Stage 1 begins at 0→10 lbs of nitrogen per acre and ends at 20→30 lbs. 8. Stage 2 begins at 20→30 lbs of nitrogen per acre and ends at 40→50 lbs. There is no Stage Stage 3 begins 50→60 lbs of nitrogen per acre and ends at 50→60 lbs. Stage 2

5.3. Marginal Product Consider feedlot stage, where cattle are fed a diet high in grain in a small pen until they are ready for slaughter. DOF = “days on feed”, the number of days the animal has been kept in a feedlot. Live-weight = the weight of the animal. Feedlots choose the DOF that maximizes profits.

TodayTomorrow DOF = days on feed LW = live-weight of steer Stage 2: the growth rate is decreasing as the calf ages—the input (DOF) is still productive, but becoming less productive. Stage 3: the cow is declining in weight—the input (DOF) hinders productivity. Stage 1: the growth rate is increasing as the calf ages— the input (DOF) is becoming more productive.

DOF = days on feed LW = live-weight of steer Without a formula for the production function, we can only measure marginal products by the rise/run between discrete points

The marginal product of Days On Feed (DOF) Cattle ID Live-weight (lbs) Days on feed (DOF)

The marginal product of Days On Feed (DOF) Cattle ID Live-weight (lbs) Days on feed (DOF)

The marginal product of Days On Feed (DOF) Cattle ID Live-weight (lbs) Days on feed (DOF)

MP = 4.36 – (DOF) MP at 25 days: MP = (25) = 3.97 MP at 200 days: MP = (200) = 1.22

MP = 4.36 – (DOF) Live- weight Days on Feed (DOF) At maximum weight the MP will equal zero. Solve for this – (DOF) = = (DOF) 4.36/ = (DOF) (DOF) = 278 at maximum weight 278

MVP = marginal value product MP tells us the additional weight of cattle from feeding them one more day. MVP tells us the value of that additional weight. MVP = (MP)(price of cattle per lb) Suppose cattle price is $0.75 per lb MVP = {4.36 – (DOF)}($0.75) MVP = 3.27 – (DOF)

Input price We feed cattle one more day whenever the MVP is greater than the cost of one DOF Suppose it costs $1.41 to feed cattle each day Price of DOF is $1.41 So we feed another day whenever MVP > $1.41

Output price = $0.75/ lb; Price of DOF = 1.41 MVP = 3.27 – (DOF) MVP at 150 DOF = 3.27 – (150) = 1.5 MVP at 150 DOF = 3.27 – (150) = 1.5 Price of one DOF is always $

Applications There are computer programs that calculate the optimal days on feed for cattle Of different breeds Of different genotypes within a breed In different environments And a targeted meat quality