Production Function, TP, MP and AP for CUET Economics

AnswerB · the technical relation between inputs and the most output they can produce

1. q = f(L, K): for each combination of labour and capital, the most output the technology allows.

AnswerD · at least one factor of production is fixed

1. In the short run a firm can change some inputs, such as labour, but not others, such as its factory.

2. In the long run every input can be varied.

AnswerA · all factors of production are variable

1. Given enough time, a firm can change even its plant and machinery.

AnswerC · The size of the factory building

1. A building cannot be enlarged at short notice; workers, materials and power can be changed with output.

AnswerB · whether all inputs can be varied, not by a fixed number of months

1. The line is drawn by flexibility, not the calendar: a tea stall's long run can be days, a steel plant's years.

AnswerC · Both I and II

1. Variable proportions: one input varies while another is fixed, so it is a short-run law. I is true.

2. Returns to scale: all inputs vary together, which is possible only in the long run. II is true.

AnswerB · a period of a year or more

1. The long run is long enough to change every input, including the shop, which takes about a year here.

AnswerC · the change in total product from employing one more unit of labour

1. MPₗ = TPₙ − TPₙ₋₁, the output the extra worker adds.

2. TP divided by labour is the average product.

AnswerA · total product ÷ units of labour

1. APₗ = TP ÷ L: output per worker.

AnswerC · 15

1. MP₃ = TP₃ − TP₂ = 39 − 24 = 15.

AnswerB · 12

1. AP₅ = TP₅ ÷ 5 = 60 ÷ 5 = 12.

AnswerD · The 8th unit, where TP is at its maximum of 70

1. MP₈ = TP₈ − TP₇ = 70 − 70 = 0: the 8th worker adds nothing.

2. TP is at its maximum (70) where MP becomes zero; after that MP is negative and TP falls to 63.

AnswerA · −7

1. MP₉ = TP₉ − TP₈ = 63 − 70 = −7.

2. Too many workers on the fixed capital get in each other's way, so output falls.

AnswerB · 18

1. TP₅ = TP₄ + MP₅ = 80 + 10 = 90.

2. AP₅ = 90 ÷ 5 = 18.

AnswerC · 30

1. TP is the sum of the marginal products: 8 + 12 + 10 = 30.

AnswerC · At 4 units of labour, where AP is at its maximum of 13

1. At 4 units, MP = 13 and AP = 52 ÷ 4 = 13.

2. MP cuts AP at the highest point of AP: before it AP rises, after it AP falls.

AnswerA · AP is rising

1. From 1 to 3 workers, MP (14, 15) is above AP, and AP rises from 10 to 13.

2. An extra worker who adds more than the average pulls the average up.

AnswerB · MP is still positive (6), so each worker still adds to output

1. TP rises whenever MP is positive, however small.

2. A falling MP only means TP rises more slowly.

AnswerD · MP is rising: 10, 14, 15

1. The slope of TP is MP. MP rising means each worker adds more than the last, so TP rises at an increasing rate.

2. With few workers on the fixed capital, specialisation and better use of the capital raise each worker's addition.

AnswerD · A is false, but R is true.

1. A is false: MP can fall while it is still above AP, and then AP does not fall. In the schedule, MP falls from 15 to 13 at the 4th worker, yet AP stays at 13.

2. R is true: what matters is whether MP is above or below AP, not whether MP is rising.

AnswerA · Both A and R are true, and R is the correct explanation of A.

1. Both are true: at 9 workers MP is −7 but AP is 63 ÷ 9 = 7.

2. R explains A: AP can only turn negative if TP does, and TP is still positive.

AnswerC · Both I and II

1. I is true: TP stops rising exactly when the extra worker adds nothing.

2. II is true: MP cuts AP from above at AP's highest point.

AnswerB · At 3 and 4 units, where AP is 13

1. AP: 10, 12, 13, 13, 12, 11, 10, 8.75, 7.

2. The highest is 13, at both 3 and 4 units; at 4 units MP also equals 13, where MP crosses AP.

AnswerC · Both I and II

1. With no variable input there is no output, so TP starts at zero: I is true.

2. MP = ΔTP ÷ ΔL, which is the slope of TP: II is true.

AnswerA · Both A and R are true, and R is the correct explanation of A.

1. Both are true, and R is the general rule behind A.

AnswerA · 6

1. MP₆ = TP₆ − TP₅ = 66 − 60 = 6.

AnswerB · 7

1. AP₉ = 63 ÷ 9 = 7.

AnswerA · Both A and R are true, and R is the correct explanation of A.

1. Both are true, and R explains A: a curve is at its peak where its slope is zero.

AnswerA · Only I

1. I is true: AP = TP ÷ L, and TP is not negative.

2. II is false: in the third phase MP is negative (−7 for the 9th worker in the schedule).

AnswerB · 8

1. TP₆ = 6 × 15 = 90; TP₇ = 7 × 14 = 98.

2. MP₇ = 98 − 90 = 8.

AnswerB · 8.75

1. TP at 8 workers is 70.

2. AP = 70 ÷ 8 = 8.75. The flat stretch means MP = 0, but AP is still positive.

AnswerC · MP is 20, 30, 25 and AP is 20, 25, 25

1. MP: 20, 50 − 20 = 30, 75 − 50 = 25.

2. AP: 20 ÷ 1 = 20, 50 ÷ 2 = 25, 75 ÷ 3 = 25.

AnswerB · 8

1. TP₆ = 6 × 18 = 108.

2. MP₆ = 108 − 100 = 8. It is below the old average (20), which is why AP fell.

AnswerA · more of one input is used with a fixed amount of another

1. Adding a variable input to a fixed one changes the proportion between them, which is why the law has its name.

2. All inputs rising together is the subject of returns to scale.

AnswerD · the law of diminishing marginal product (returns to a factor)

1. Its central claim is that, after a point, each extra unit of the variable input adds less: diminishing marginal product.

AnswerB · TP rises at an increasing rate and MP rises

1. At first the fixed input is under-used, so adding workers raises output more than in proportion.

2. In the schedule, MP rises from 10 to 15 over the first 3 workers.

AnswerC · TP rises at a diminishing rate while MP falls but stays positive

1. Each extra worker now adds less than the one before, but still adds something.

2. The phase ends where MP reaches zero and TP is at its maximum.

AnswerC · MP is negative and TP falls

1. Too much of the variable input crowds the fixed input, so an extra unit lowers output.

2. No rational firm hires into this phase.

AnswerD · Each worker has less of the fixed input to work with

1. The fixed input is shared among more and more workers.

2. Once it is fully used, each extra worker adds less: the proportion between inputs has moved past its best.

AnswerA · From the 1st to the 3rd unit

1. MP: 10, 14, 15 for units 1 to 3, then 13, 8, 6, 4, 0, −7.

2. MP rises only up to the 3rd unit; from the 4th it falls.

AnswerA · Both A and R are true, and R is the correct explanation of A.

1. Both are true.

2. If every input rises together, their proportion does not change, so there is nothing for the law to describe: R explains A.

AnswerA · Only I

1. I is true: MP is positive, so TP rises.

2. II is false as a rule: AP rises only while MP is above it; in the table AP holds at 13 and then falls from the 5th worker, all within the second phase.

AnswerA · A-II, B-IV, C-I, D-III

1. MP is the slope of TP: rising MP, steepening TP; falling but positive MP, flattening but rising TP.

2. Zero MP is the top of TP; negative MP means TP falls.

AnswerD · A is false, but R is true.

1. A is false: the workers are assumed equally skilled; MP falls because the fixed input is spread thinner.

2. R is true and is why skill cannot be the reason.

AnswerA · diminishing marginal product

1. One input (fertiliser) rises while another (land) is fixed, and each extra unit adds less.

AnswerA · More workers on a fixed number of machines

1. The law needs one input varied while another stays fixed.

AnswerA · Only I

1. I is true: in the first phase MP is rising and above AP, so AP rises.

2. II is false: in the third phase MP is negative, far below AP, so AP falls.

AnswerB · The second phase, where MP falls but is positive

1. In the first phase each extra worker adds more than the last, so it pays to keep hiring.

2. In the third phase extra workers lower output. So the firm settles in the second phase.

AnswerD · The 9th unit, where MP becomes negative

1. The third phase is where TP falls, i.e. MP is negative.

2. TP falls from 70 to 63 at the 9th unit (MP = −7).

AnswerC · all inputs are increased in the same proportion

1. Returns to scale compare the rise in output with the proportional rise in every input together, a long-run idea.

AnswerB · increasing returns to scale

1. Output rising by a larger proportion than inputs is increasing returns to scale.

AnswerB · constant returns to scale

1. Output rising in exactly the same proportion as inputs is constant returns to scale.

AnswerA · decreasing returns to scale

1. Doubling inputs would give 200 under constant returns; 170 is less than double.

2. Output rising by a smaller proportion than inputs is decreasing returns to scale.

AnswerA · Only I

1. I is true: all inputs can vary together only in the long run.

2. II is false: under returns to scale no factor is fixed.

AnswerD · Greater division of labour and specialisation in a larger firm

1. A bigger firm can split work into specialised tasks and use larger, more efficient machines.

2. Management difficulties are a reason for decreasing returns to scale.

AnswerB · Problems of co-ordinating and supervising a very large organisation

1. As a firm grows very large, management becomes harder and output may rise less than inputs.

2. The other three are sources of increasing returns.

AnswerD · A-III, B-I, C-IV, D-II

1. 250 is more than double: increasing; 200 is exactly double: constant; 160 is less: decreasing.

2. If only labour doubles, it is a question of returns to a factor, not of scale.

AnswerA · Both A and R are true, and R is the correct explanation of A.

1. Both are true, and R explains A: cost and output rise in the same proportion, so cost per unit is unchanged.

AnswerD · doubles output: constant returns to scale

1. (2L)^0.5 (2K)^0.5 = 2^(0.5 + 0.5) L^0.5 K^0.5 = 2q.

2. The exponents add to 1, so output doubles: constant returns to scale.

AnswerC · increasing returns to scale, since 0.6 + 0.6 > 1

1. Scaling both inputs by t multiplies output by t^(0.6 + 0.6) = t^1.2.

2. t^1.2 > t for t > 1, so output rises more than in proportion.

AnswerA · Both A and R are true, and R is the correct explanation of A.

1. Both are true: q = L^0.5 K^0.5 has diminishing MP of labour alone but constant returns when both double.

2. R explains A: one law varies one input, the other varies all, so they need not agree.

AnswerC · raises output by less than double: decreasing returns to scale

1. The exponents add to 0.3 + 0.5 = 0.8, less than 1.

2. Output rises by 2^0.8 ≈ 1.74 times: less than double. The constant 2 in front does not affect returns to scale.