(4,10) (8,14) sometimes we can’t observe the production function in green. but we can observe how...

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(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

Yield

5.3. Marginal Product


Produce

Wheat?

Change in N (lbs / acre)

Change in Wheat

yield (bushels / acre) MP

NO ----- ----- -----Yes 0 →10 23→25 0.20 Yes 10→20 25→30.5 0.55 Yes 20→30 30.5→33 0.25 Yes 30→40 33→35 0.20 Yes 40→50 35→36 0.10 Yes 50→60 36→36.2 0.02

MP = rise/run

Going from 0 to 10 lbs N:

MP = (25-23)/(10-0)

= 2/10= 0.2


Produce Wheat?


Change in Wheat yield

(bushels / acre) MP

NO ----- ----- -----Yes 0 →10 23→25 0.20 Yes 10→20 25→30.5 0.55 Yes 20→30 30.5→33 0.25 Yes 30→40 33→35 0.20 Yes 40→50 35→36 0.10 Yes 50→60 36→36.2 0.02

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.


Produce Wheat?



(bushels / acre) MP

NO ----- ----- -----Yes 0 →10 23→25 0.20 Yes 10→20 25→30.5 0.55 Yes 20→30 30.5→40 0.95 Yes 30→40 40→35 -0.50 Yes 40→50 35→33 -0.20 Yes 50→60 33→29 -0.40

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 2.6. 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.


Produce Wheat?



(bushels / acre) MP

NO ----- ----- -----Yes 0 →10 23→25 0.20 Yes 10→20 25→30.5 0.55 Yes 20→30 30.5→40 0.95 Yes 30→40 40→45 0.50 Yes 40→50 45→47 0.20 Yes 50→60 47→43 -0.40

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 2.9. Stage 3 begins 50→60 lbs of nitrogen per acre and ends at 50→60 lbs.

Stage 2


• 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.

Today

Tomorrow

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

IDLive-

weight (lbs)

Days on feed (DOF)

12408 601 1

12408 1240 138

12408 1398 180

12411 608 1

12411 1210 138

12411 1316 180

Marginal Product

(MP) of DOF between 1 and 138 DOF is

=

=

=4.66 lbs per extra

DOF


IDLive-

weight (lbs)

Days on feed (DOF)

12408 601 1

12408 1240 138

12408 1398 180

12411 608 1

12411 1210 138

12411 1316 180

Marginal Product

(MP) of DOF between 138

and 180 DOF is

=

=

=3.76 lbs per extra

DOF


IDLive-

weight (lbs)

Days on feed (DOF)

12408 601 1

12408 1240 138

12408 1398 180

12411 608 1

12411 1210 138

12411 1316 180

MP between 1 and 138

DOF=

=4.39 lbs per extra

DOFMP between 138 and 180

DOF=

=2.52 lbs per extra

DOF

MP = 4.36 – 0.0157(DOF)

MP at 25 days: MP = 4.36-0.0157(25) = 3.97MP at 200 days: MP = 4.36-0.0157(200) = 1.22

MP = 4.36 – 0.0157(DOF)

Live-weigh

t

Days on Feed (DOF)

At maximum weight the MP will equal zero. Solve for this.4.36 – 0.0157(DOF) = 04.36 = 0.0157(DOF)4.36/0.0157 = (DOF)(DOF) = 278 at maximum weight

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 – 0.0157(DOF)}($0.75)• MVP = 3.27 – 0.011775(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

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2

2.1

2.2

2.3

2.4

2.5

2.6

2.7

2.8

2.9

3

3.1

3.2

3.3

3.4

0 50 100 150 200 250

$ p

er D

OF

Days on feed (DOF)

Set MVPDOF = PDOF

(MP)($0.75) = $1.41{4.36 – 0.0157(DOF)}

($0.75) = $1.41Profit maximizing DOF =

= DOF

MVP = 3.27 – 0.011775(DOF)

MVP at 150 DOF = 3.27 – 0.011775(150) = 1.5

Price of one DOF is always

$1.41

ApplicationsThere 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

(4,10) (8,14) sometimes we can’t observe the production function in green. but we can observe how...

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