prod func part - 1

8/9/2019 Prod Func Part - 1

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Production Functions

The major goal of production economics is to increase the efficiency of agriculturalproduction. This esssentially entails to answer the following questions :

1] How agricultural production can be managed efficiently ? i.e. Sustained growthof agricultural production.

2] How do the inputs combinations minimize the cost on sustained basis ?3] How will output respond to change in input output prices ? e.g. Whether the

withdrawal of fertilizer subsidy would affect the output ?4] What is the combination of enterprises that will maximize our profit ? e.g. Different

crops + Dairy + Poultry, etc.5] How will agricultural production be affected by various levels of technology ?6] How do you plan under uncertain situations ? e.g. Drought, flood, etc.

The production economics includes intra and inter farm allocation of agriculturalresources during a given period as well as over a period of time. It treats generalprinciples for allocation of land, labour, capital and management inputs, which havescarced amount and alternative uses, so as to achieve a predefined objective such asprofit maximization, etc.

Production Function : The relationship between inputs and outputs can be characterized as a production

function. Production function is thus a technical and mathematical relationshipdescribing the manner and extent to which a particular product depends upon themagnitude of inputs used. In production function, output is "dependent upon" or "relatedto" or "determined by" or is the "function of" inputs.

Now a days, inputs, factors and resources are inter-changeably used. But there

exists subtle distinction among them, e.g. the word 'input' is used for seed, fertilizer,etc. the word "resources" is used for natural resources i.e. land, while the "factor"includes both input as well as resource.

Wicksteed used the production function for the first time in 1894, but Wicksell became one of the first economists to treat the production function explicitly in the year 1901. The production function for multi-product, multi-factor form was developed by Hicks in 1939. Relatively more recent works were done by the economists like

Carlson, Dano, Frisch, Monger and Samuelson.


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Assumptions for Estimating Production Function :

1. Perfect Knowledge : We should have perfect knowledge of prices of input andoutput, availability of resources, markets, etc. for maximizing profit.

2. The values of inputs and output should be non-negative i.e. Y ≥ 0 and Xi ≥ 0(i = 1, 2, ...., k)

3. Homogeneity of product and input : The production function is defined only for the same group i.e. observation should not be taken from different groups.e.g. few observations of groundnut crop and few observations of bajra crop shouldnot be merged together for estimating a production function. Similarly, theproduction functions for crossbred cow, desi cow and buffalo should be estimatedseparately.

4. All the factors and products are perfectly divisible.

5. Length of time : The length of time should be sufficiently short that we are (i)unable to alter the fixed resources and (ii) unable to change the technology, but it should be large enough that the process of transformation is completed.

6. Technology : The technology (biological or mechanical) is assumed to remainsame for the group under study.

7. The production function presupposes technical efficiency. This means that every possible combination of input is assumed to result in maximum level of output.

8. The production function / input-output relationship is single valued andcontinuous. i.e. values should not be in range.

9. Fixed factors are held constant.

10. Consumers' tastes and preferences and other factors influencing consumers'demand are given.

11. There are no institutional constraints for using the resources.

12.Returns to scale is assumed to be decreasing.

≥ ≤


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13. The production function is characterized by declining marginalproduct for all factor-product combinations.

14. Rate of product transformation between any two products is assumed

to be increasing.

The production function can be expressed in the following ways :(i) Arithmetically (ii) Geometrically or graphically and(iii) Algebraically

The graphical presentation of classical / cubic / transcendentalproduction function is explained below :

Y

X ENO

A

B C

D

F

II nd ZoneRational

Zone1 > EP > 0

III rd ZoneIrrational

ZoneEP < 0

EP > 1

EP = 1First ZoneIrrationalZone

EP = 0 T. P. Curve

A. P. Curve

M. P. CurveInput

O u t p u t

Fig. 1 : Three stages of production function

This production function describes three stages of production, hence,it is called three zones / stages of production.

Zone - I (Stage - I) : The first zone starts from the origin and ends where the marginal

product (MP) = Average product (A.P.). In the diagram, at the point 'C',M.P. = A.P. and hence the elasticity of production (EP) = 1 ( EP = M.P. /

A.P.). Thus 'ON' portion is called first zone. The point 'A' in the total∴


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It means it is the point where the T.P. curve changes its curvature and the M.P. will bethe highest i.e. point 'B' on M.P. curve. In the first zone, T.P. is increasing at an increasingrate upto the point of inflection, because M.P. is increassing upto that point. Beyondthe inflection point, T.P. is increasing at decreasing rate. M.P. curve lies above the A.P.

curve throughout the first zone, indicating that M.P. > A.P. The marginal product andaverage product will be equal at the end of first zone i.e. point 'C' in the diagram. The A.P. is maximum at point 'C'. This implies the unit elasticity of production. The first zone is called irrational zone because the production is increasing at increasing rate inthis zone. The elasticity of production is greater than one indicating that, one percent increase in the level of input, results into more than one per cent increase in the output.

This suggests that the resources are under utilized and, if more resources are used, theproduction / output will increase. It can thus be inferred that the optimum level of output is not achieved in the first zone. The probable reasons for under utilization of resources could be the lack of knowledge about optimum level of output, non-availability of inputs, indivisibility of resources e.g. labour upto optimum level,inadequate capital, inefficient techniques of farming, uncertainty of future yield andprice, lack of alternative, laziness, etc. to overcome such promblems at marco level,the extension agencies and government policies will play vital role.

Zone - II (Stage - II) :'NE' portion in the diagram is called second zone of production. The second zone

starts from the point where MP = AP or AP is maximum i.e. point 'C' and ends at thepoint where MP becomes zero i.e. point 'E'. The range of elasticity is zero to one in thiszone, indicating that the total production is increasing but at decreasing rate. The TPreaches at maximum, when MP becomes zero, i.e. EP also becomes zero. The AP startsdeclining but the AP curve lies above the MP curve throughout the zone. This zone iscalled rational zone, because the resources are utilized rationally / optimumly in thiszone and the producer gets maximum output in this zone only. After this zone, the totalproduction starts declining because MP and EP become negative.

Zone - III (Stage - III) : The third zone starts after the end of second zone. Total production starts declining

in this zone. Both MP and EP become negative. This region is called irrational becauseresources are over utilized here. It means the withdrawal of certain quantum of input

will not adversely affect the output, or the same output can be produced with lessmagnitude of input. This over utilization of resources could be attributed mainly to theignorance of producers, lack of technical know-how, very low prices of inputs, etc.

Problem Exercise :Q.1. What does the inflection point indicate ?


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Ans. Third zone, because negative MP is found in the third zone only.

Q.13. What is fabrication coefficient ? Ans. It means inverse relationship of AP i.e. X/Y; this shows that how much input is

required to produce one unit of output e.g. 500 kg paddy is required to produce50 quintals of paddy, then 500/50 = 10. This is the fabrication coefficient.

Q.14. Why is the third zone called irrational ? Ans. Because of over utilization of resources, i.e. more quantities of resources are

used than actually needed.

Q.15. If the maximum value of AP is 15 units. Find out the values of MP and EP. Ans. The value of MP will also be 15 units because at maximum AP, both MP and AP

will be equal (at the end of first zone), hence EP will be one ( EP = MP/AP).

Q.16. In which zones, the AP curve lies above the MP curve ? Ans. In the second and third zone.

Q.17. What happens to AP, EP and MP, when TP declines ? Ans. When TP declines, the AP also declines and both MP as well as EP become

negative.

Q.18. In which zone, the maximum MP is found ? Ans. In the first zone.

Q.19. What will be the value of EP at the point of inflection ? Ans. Greater than one.

Q.20. If MP = 3 units, AP = 4 units, what will be the value of EP ? What is its

interpretation? Ans. EP = MP/AP = 3/4 = 0.75. This implies that one per cent increase in the level

of input will result into 0.75 percent increase in the output.

Q.21. What happens to AP and EP, when MP is zero ? Ans. EP will also be zero, but AP will be positive.

Q.22. In which zone, TP first increases at increasing rate and then at decreasing rate? Ans. In the first zone, upto the point of inflection, TP increases at increasing rate

and then increases at declining rate.

∴


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Q.23. In the beginning of second zone, if TP of 20 units is obtained by using 5 unitsof input, find out the values of MP and EP.

Ans. AP = = = 4 units.

In the beginning of second zone, both AP and MP will be same, hence MP = 4units, and therefore EP = 1 unit.

Q.24. In which zone, the declining trend of MP but increasing trend of AP is observed? Ans. In the first zone, after the point of inflection, MP starts declining, but AP goes

on increasing.

Q.25. If TP is zero, what will be the value of AP ? Ans. AP will also be zero.

Q.26. When MP is increasing, how does AP behave ? Ans. When MP is increasing, the AP is also increasing, but its value will be less than

MP. (In the first zone.)

Q.27. If 25 kg of fertilizer dose is applied, the farmer will obtain maximum groundnut production of 1500 kg per hectare. What will be the elasticity of production at

that point ? Ans. The elasticity of production will be zero, because 1500 kg output is the

maximum production of groundnut, hence, MP at that point will be zero, andEP = MP/AP = 0/AP = 0.

* A Summary of Relationship :

Question Answer

A] What happens to TP ?

i) When MP is increasing TP is increasing at increasing rate.ii) When MP is decreasing but it is

greater than zero TP increases at decreasing rate.iii) When MP = 0 TP is maximum.iv) When MP is negative TP is decreasing.

v) When AP is increasing TP first increases at increasing rate(i.e. upto the point of inflection) andthen at decreasing rate.

Total production (Y)Quantity of input (X)

205


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Question Answer

B] What happens to AP ?

i) When MP > AP AP is increasing.ii) When MP = AP AP is maximum.iii) When MP < AP AP is decreasing.iv) When MP increases AP also increases.

v) When TP is zero AP is zero.

Summary of Zone Comparison :

Variable Zone I Zone II Zone III Total production It increases first Increases at decreasing Decreases

at increasing rate ratethen at decreasingrate

Marginal product First increases Decreases but Negativethen decreases remains positive

Average product Increases conti- Decreases but Decreases but nuously but it is more than positiveremains less than marginal product marginal product

Elasticity of More than unity Less than one but Negativeproduction greater than zero

Terminologies used in Production Economics :1. Product : It is the result of the use of resources or services of resources, e.g.

wheat, milk, maize, etc.2. Production : It is the process of transformation of resources / inputs, like seed,

fertilizers, etc. into products.

3. Average product : It is ratio of total product to the quantity of input used inproducing that amount of product.

AP =

4. Average marginal product :It refers to the quantity which additional unit of factor / input adds to the

total product. MP = i.e. it is the rate of change in total product as the quantity of input changes.

T.P. (Y)Quantity of input (X)

∆ y ∆ x


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5. Exact marginal product :It means marginal product at a point, indicated as a derivative.

e.g. For function Y = a + bx,

MP = = b

6. Choice indicator :It is a yardstick or an index or a criterion indicating which of the two or

more alternatives is optimum or will maximize a given end, e.g. price ratio &substitution ratio, MR = MC, etc.

7. Slope of curve :It is the rate of change in one quantity in response to the change in the other.

∂ y ∂ x

Y

X 10

51015

Input

O u t p u t

∴ Solpe of curve =

= = 0.5

It means one unit of input will add 0.5unit of output.

∆ y ∆ x 5

10

%∆ y %∆ x

∆ y y

∆ x x

8. Elasticity of production :It refers to the percentage change in output in response to the percentage

change in input.

i.e. EP = = ( * 100) ÷ ( * 100)

=

∴ EP = ( ∆ y/ ∆ x = MP, and 1/AP = x/y)

9. Isoquant or Iso-product curve : (Iso = same, Quant = production)It represents the various input combinations that can be used to produce a

given output.

x

y

∆ y

∆ x MP

AP

20

∴


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Ridgeline for X 1

Ridgeline for X 2

y = 120 kg.

y = 105 kg.

y = 80 kg. y = 50 kg.

IsoclineScale line /Expansion path

0

Isoquant

Von Leibig point < <

<

< < <

N.B.: Any isocline joining least cost combination points is called expansion path.It may or may not pass through origin.

16. Asymptotic curve : The curve which does not touch to 'X' axis is called asymptotic curve, e.g.

Isoquant.

17. Marginal rate of substitution :It is the rate at which one input is substituted for another.

e.g. MRS of X 1 for X 2 =

18. Elasticity of substitution (ES) : The concept was first given by Robinson and H.R. Hicks in 1930.

Proportionate change in the input ratio due to proportionate change in the rate of technical substitution, keeping the output level at constant, is called elasticity of substitution.

ES of x 2 for x 1 =

=

19. Marginal rate of product substitution (MRPS) :It is the substitution ratio for two products, while MRS or MRTS is for factor

substitution.

∆ X 2∆ X 1

% ∆ (x 1 / x 2)% ∆ (Rate of Technical substitution)

d (x 1 / x 2)d (dx 1/dx 2)

(dx 1/dx 2)(x 1 / x 2)*

X 2

X 1


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MRPS of y 2 for y 1 = =

∴ MRPS of y 2 for y 1 =

Because MPP of y 1 =

MPP of y 2 =

Now MRPS of y 2 for y 1 = (MPP of y 2) ÷ MPP of y 1

= ÷

MRPS of y 2 for y 1 =

∆ y 1∆ y 2

dy 1dy 2

MPP of y 2MPP of y 1

dy dy 1dy dy 2

dy dy 2

dy dy 1

dy 1dy 2

Types of Production Functions :

1. Linear function : The form of the function is : Y = a + bx The values of 'a' and 'b' may be positive or negative. Here 'a' =

constant, i.e. if the value of 'x' = 0, then y = 'a' i.e. some constan©©©t or fixed production, which may be due to some other factors not included in the function.

Minimum requirement of resources -a

0

a

k

Y

X

y = a +

b x

y = - a +

b x

In case of y = -a + bx, if x = 0, then y = -a, i.e. at least 'k' resources are required for even negligible production, e.g. minimum moisture in the soil is required to produce at least some production. We are interested to get at least some output and not interestedin zero output, thus, some resources are required.

>


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Y

X 650

1

-a

kg.

e.g. Here, to get 1 kg. of output, at least 6kg. of X is required.

General form of the function : y = a + b 1 x 1 + b 2 x 2 + ............................. + b n x n,Here 'b' indicates that the output increases by this rate by changing one unit of

resource.

Characteristics :1. Marginal physical product (MPP) :

MPP = = bi ≠ 0,dy dxi

Y

X O

a O u t p u t

We do not get maximum output in linear function, because MPP i.e. 'b' never tendsto zero, but it is constant.

(dy/dx 1)(dy/dx 2)

MP x 1MPx 2

bc

2. At each stage, EP is different. This is the benefit of the linear function.EP = MP / AP = (b) (x/y). Here 'b' i.e. MP is constant, but APP is changing and

thus, EP will also be changed.

3. MRTS :For y = a + bx 1 + cx 2

MRTS of x 1 for x 2 = = =

OR = = = constant = k,

because 'b' and 'c' are constnat. So, at each level of output, MRTS would be same.

d x 2dx 1

bc

Input


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−

− y - a

bc

b( ( ( (

− y - a b

e.g. y = 10.50 + 3x 1 + 7.5x 2 ∴ MRTS = b/c = 3/7.5 = 1/2.5 = 0.40

Thus, for every stage of production, MRTS will be 0.40 => MRTS is a linear function without intercept in linear production function.

4. Isoquant : y = a + bx 1 + cx 2For isoquant, we have to get the values of x 1 or x 2.

∴ bx 1 = y - a - cx 2, → 'y' is constant because isoquant is always for a given levelof output.

Solve for x 1 :

x 1 = x 2 ,

since 'y' and 'a' have some fixed values, they are written together.

Putting = ∝ and c/b = β ,

∴ x 1 = ∝ - β x 2. This implies that an isoquant is a linear function.

Properties of isoquant :

i) Linear :

ii) Negatively sloped X 2

O

iii) Decreasing, but two isoquantsdo not intersect each other. y = 80 kg.

y = 50 kg.

X 2

X 1

Oiv) Constant slope :

i.e. change in input level is same in every stage.

X 2O

X 1 Here AB = BC = CO, ∴ x 1 = ∝ - β x 2, if x 2 = 0 then x 1 = ∝, This shows the perfect substitution. It means by withdrawingurea(x 2) completely, the yield will beconstant, by increasing another input (x 1)i.e. F.Y.M.

A

BC

F.Y.M.

Urea

X 1

Assumptions of Isoquant:

1) If you add one input, you have to reduceother input, to maintain same level of output.2) If addition and substraction of input is one,the MRS=-1, i.e. constant. It is observed inlinear production function.3) The rational isoquant is one with decreasingMRS.4) If MRS=0, the production function is y=a.


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v) Convex to the origin and decreasing MRTS.

vi) If one isoquant lies above and to the right of another isoquant, the former willshow higher output than the latter.

X 1

X 2O

y 2 = 125 kg y 2 = 80 kg

5. Isocline := = = = k,

This is possible only if b 1 and b 2 i.e. Marginal products are constant and which is practically not possible. This shows that we do not get isocline in a linear function.

6. Ridgelines :i.e. the slope of isoquant should be zero. It is important to determine the

zones of production.= = 0 ; But ≠ 0,

Thus, in a linear function, we can not identify the ridgelines, hence rationalzone is also not possible; i.e. we cannot estimate the maximum profit or optimumlevel of input.

dy/dx 1dy/dx 2

dx 2dx 1

MP of x 1MP of x 2

b1 b2

dx 1dx 2

b1 b2

b1 b2

O

Y

X

No rational zone Rational zone

O

Y

X 7. Linear function gives directly marginal product or marginal value product.8. Elasticity of substitution is infinity.9. Economic optima under a budget constraint can be defined as

=

Example 1 : Fit a linear production function assuming production as y, and x 1 and x 2 as inputs using the data given in Table 1.

y = a + b 1 x 1 y = a + b 1 x 1 + b 2 x 2.

b 1 b 2

P x 1 i.e. Ratio of MPs is quated to the ratio of input prices.P x 2


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Sr.No. Y X 1 X 2 y=Y-Y x 1=X 1-X x 2=X 2-X 2

1 3.16 2.96 2.29 -1.904 -1.636 0.6152 5.49 5.19 2.34 0.4259 0.594 0.6653 9.60 8.12 1.96 4.536 4.324 0.2854 6.50 3.82 2.14 1.436 -0.776 0.4655 3.68 3.55 0.90 -1.384 -1.046 -0.7756 1.24 3.0 1.68 -3.824 -1.596 .0057 3.44 2.47 1.42 -1.624 -2.126 -.2558 6.70 7.43 1.51 1.666 3.834 -0.1659 3.86 2.91 1.02 -1.204 -1.686 0.655

10 10.33 6.31 2.27 5.266 1.714 0.59511 5.03 5.01 1.53 -0.034 .414 -0.145

12 1.89 3.59 1.05 -3.174 -1.006 -0.625

∑ x 12 = 39.15

∑ x 22 = 3.005 y = 5.0767

∑ x 1 y = 48.12 x 1 = 4.53∑ x 2 y = 8.18 x 2 = 1.675∑ x 1 x 2 = 3.60

−

(1) Estimate the values of a & b 1 and b 2(2) Estimate the Standard Error (S.E.) of b 1 & b 2(3) Work out the coefficient of multiple determination (R 2)

For the single variable y = a + b 1 x 1,

b 1 = = = 1.229

a = y - b 1 x 1 = (5.0767) - (1.239 * 4.53)

∴ a = - 0.491

R 2 = = = 0.67

48.1239.15

− −

b1∑ x 1 y ∑ y 2

(1.239) x (51.826)87.651

Thi S shows that about 67 per cent variation is explained by x 1 input only.σ 2 = ∑e2 / (n - k - 1) where k = total no. of independent variables

n = total no. of observations

∑ x 1 y ∑ x 1

2

− − −

∑ 60.92 54.36 20.11


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σ 2 x 1∑ x 2

2.6045.485

σ 2 x ∑ x 2

n∑ x 2

∑e2 = ∑ y 2 - b 1 ∑ x 1 y = 87.651 - 1.139 x 51.826= 28.621

∴ σ 2 = 28.621 / (11)

= 2.60

S.E. =

=

= 0.239 y = - 0.08297 + 1.139x 1

S.E. =

y = a + b 1 x 1 + b 2 x 2

b 1 =

=

= 1.0819

This shows that one unit increase in the level of x 1 will result into 1.099 unitsincrease in output.

t = = (1.0189) / (0.2284) = 4.74 i.e. highly significant.

=

=

= 1.6117

^

(51.826 x 3.005) - (6.598 x 1.62178)(45.485 x 3.005) - (1.62178) 2

b1

SE ( )

∑ x 2 y ∑ x 12 - ∑ x 1 y ∑ x 1 x 2

∑ x 12 ∑ x 2

2 - ( ∑ x 1 x 2)2 b2

^

(6.598 x 45.485) - (51.826 x 1.6217)134.0525

b2^

∑ x 1 y ∑ x 22 - ∑ x 2 y ∑ x 1 x 2

∑ x 12

∑ x 22

- ( ∑ x 1 x 2)2

b1^


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Ex.2. The input output data for 12 farmers are given in the following table.

Sr. No. X 1 X 2 Y

1 2.96 2.29 3.16

2 5.19 2.34 5.193 8.92 1.96 9.604 3.82 2.14 6.505 3.55 0.90 3.686 3.00 1.68 1.247 2.47 1.42 3.448 7.43 1.51 6.739 2.91 1.02 3.86

10 6.31 2.27 10.3311 5.01 1.53 5.0312 3.59 1.05 1.89∑ 55.16 20.11 60.65

From the above data, estimate the followings :1. Linear Y = a + b 1 x 1 + b 2 x 22. Standard errors of the regression coefficients, SE(b i) and 't' values.

3. Coefficient of multiple determination (R 2

).4. Correlation coefficients. 5. MP. 6. RTS. 7. Isoquant.

Step I : Formulate the normal equations.Step II : Work out the summations, sum of square, and sum of products from the raw

data Step III : Work out the corrected values.Step IV : Work out the regression parameters.

Step V : Estimate the standard errors of regression coefficient.Step VI : Estimate the 't' value of regression coefficients.Step VII : Estimate the coefficient of multiple determination.Step VIII : Work out the marginal products.Step IX : Work out the R.T.S.Step X : Work out the Isoquant equation.

Interpret the results

^


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STEPS : A. Formulate normal equations :

∑ y = na + b 1 ∑ x 1 + b 2 ∑ x 2∑ x 1 y = a ∑ x 1 + b 1 ∑ x 1

2 + b 2 ∑ x 1 x 2

∑ x 2 y = a ∑ x 2 + b 1 ∑ x 1 x 2 + b 2 ∑ x 22

B. Work out the following values :i] ∑ x 1 = 55.16 vii] ∑ x 1

2 = 299.04ii] ∑ x 2 = 20.11 viii] ∑ x 2

2 = 36.71iii] ∑ y = 60.65 ix] ∑ Y 2 = 405.8533iv] X 1 = 4.5967 x] ∑ x 1 x 2 = 96.2694

v] X 2 = 1.6758 xi] ∑ x 1 y = 330.4367 vi] Y = 5.0542 xii] ∑ x 2 y = 109.616(These values are to be worked out with the help of raw data)

C. Worked out the following corrected values :

i] ∑ x 12 = ∑ x 1

2 - = 45.48787

ii] ∑ x 22 = ∑ x 2

2 - = 3.00899

iii] ∑ y 2 = ∑ y 2 - = 99.3181

iv] ∑ x 1 y= ∑ x 1 y - = 51.64887

v] ∑ X 2 Y= ∑ x 2 y - = 7.97671

vi] ∑ X 1 X 2 = ∑ x 1 x 2 - = 3.8304

Where n = no. of observations.

D. Work out the regression parameters :

i] D = ∑ x 12 ∑ x 2

2 - (∑ x 1 x 2)2

= 122.19675ii] b1 = (∑ x 1 y ∑ x 2

2 - ∑ x 2 y ∑ x 1 x 2) / D = 1.0217iii] b2 = (∑ x 2 y ∑ x 1

2 - ∑ x 1 y ∑ x 1 x 2) / D = 1.3501iv] a = y - b 1 x 1 - b 2 x 2 = - 1.9047

( ∑ x 1 ) 2

n( ∑ x 2 )

2

n( ∑ y) 2

n ∑ x 1 ∑ y

n ∑ x 2 ∑ y

n ∑ x 1 ∑ x 2

n

− − −

− −

−


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E. The standard error estimation :i] ∑e2 = ∑ y 2 - b 1 ∑ x 1 y - b 2 ∑ x 2 y = 35.778994ii] σ 2 = ∑e2 /(n - k - 1) = 3.9754

Where k = No. of independent variables

iii] Variance (b 1) = = 0.09789

iv] Variance (b 2) = = 1.47984

v] SE (b 1) = Variance (b 1) = 0.31287

vi] SE (b 2) = Variance (b 2) = 1.21647

F. Estimate 't' values :i] 't' ratio for a) b 1 = = 3.2655

b) b 2 = = 1.1098

ii] Compare calculated 't' values with theoretical values at (n - k - 1) d.f. at 5per cent probability level.i) b 1 is significant ii) b 2 is non significant

G. Calculate coefficient of multiple determination :

R 2 = = 0.63975

H. Correlation coefficient between independent variables :

rx 1 x 2 = = 0.327457

I. Interpretation of results :

i] MP of x 1 = = 1.0217, Similarly, MP of x 2 = = 1.3501

ii] RTS = = = 0.7568

iii] Isoquant equations :e.g. for x 1 :- Y = a + b 1 x 1 + b 2 x 2

∴ b 1 x 1 = y - a - b 2 x 21.0217 x 1 = (5.0542 + 1.9047) - 1.3501 x 21.0217x 1 = 6.9589 - 1.3501 x 2 , x 1= 6.8111 - 1.3214x 2

σ 2 ∑ x 22D

σ 2 ∑ x 21D

b1SE(b 1)

b2SE(b 2)

b 1 ∑ x 1 y + b 2 ∑ x 2 y ∑ y 2

∑ x 1 x 2∑ x

1

2 ∑ x 2

2

dy dx 1

dy dx 2

MP of x 1MP of x 2

1.02171.3501

−


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2) Cobb. Douglas Production Function :

Cobb and Douglas first applied this function for empirical estimation using thetime series data pertaining to the American manufacturing industries over the period1899 to 1922. This function is also known as Power function, log linear function, loglog function or double log function. The algebraic forms of the function with one, andmore than one variables are given below :

Original form Log form Y = ax b Log y = log a + b log x Y = ax 1

b1 x 2 b2 Log y = log a + b 1 log x 1 + b 2 log x 2

Y = ax 1 b1 x 2

b2 .......... x n bn Log y = log a + b 1 log x 1 + b 2 log x 2 + .... + b n log x n

∴ General form is :

Y = a Xi bi Log Y = log a + b i log x i

The graphic form is as under :

πn

i = 1

n

i = 1∑

X

Y

Y

Y

X

X

b > 1

b = 1

O

O

dy dx 1

Fig. : Graphic illustrations of power function with one input variable.

Characteristics :1. Marginal product :

y = ax 1 b1 . x 2

b2

∴ MP of x 1 = = b 1 ax 1 b1-1 . x 2

b2

= ( ∴ x 1 b1-1

= )

But the encircled portion i.e. ax 1 b1 . x 2

b2 = Y,

. x 2

b2ax 1 b1

x 1

x 1 b1

x 1

when b < 1

y = f (x 1)

O

∴ x 1 = - x 2 = 6.8111 - 1.3214x 26.95891.0217

1.35011.0217


24/55

∴ MP of x 1 = =

Now, for single variable function (y = ax b) :

i] If b = 1, then,

dy/dx = b . ax b-1 , Since b = 1, x 1-1 = x 0 = 1∴ MP of x = dy/dx = a

iii] If b < 1dy/dx = (ab . x b )/x => MP declines, e.g. b = 0.5 and x = 4, i.e. Y = ax b = ax 0.5

then dy/dx = 0.5 a . x -0.5 = 0.5a = 0.5a

∴ dy/dx = . =Now if x = 9, and b = 0.5,

∴ dy/dx = = a/6

2. When b = 1, this function becomes a linear function. ∴ Y = ax b , if b = 1, it becomes y = ax.

3. In Cobb Douglas function, we get either constant, increasing or decreasing returns/MP, but not all the three of even any two at a time.

4. Elasticity of production (EP) is constant :

EP = MP/AP, or MP =(EP) (AP).e.g. Y = ax b , ∴ dy/dx = b . ax b / x = b.y/x, here y/x = AP.

Now, EP = MP/AP, but MP = b.y/x, and AP = y/x ∴ EP = . = b

Thus, the EP is constant in Cobb Douglas function, or this function directly givesus the elasticity of production.

5. Elasticity of Substitution (ES or θ) is one

ES or θ =

dy dx 1

b 1.y x 1

x 4

0.5a

9

Proportionate ∆ in Input ratioProportionate ∆ in Rate of TechnicalSubstitution (RTS)

b = 1

X O

Y

MP

ii] If b > 1MP = dy/dx = a.bx b /x => MP increases e.g. if b = 2,∴ dy/dx = 2a x 2 / x = 2ax, Here if x = 2, then MP

becomes 4a, and if x = 4, it becomes 8a. i.e. MP increases because

of greater than one value of b.

MP

b > 1 Y

X

MP

Y

X

MP < 112

a 4

a 4

b.y x

x y


25/55

∴ θ of x 1 for x 2 =

∴θ = *

Now in C.D. function : y = ax 1

b1 - x 2 b2

dy/dx = b 1 . a x 1 b1 x 2

b2 This is 'Y' x 1

MP of x 1 = dy/dx 1 = b 1 . y / x 1Similarly, MP of x 2 = dy/dx 2 = b 2 y / x 2dividing MP of x 1 by MP of x 2 :

=

∴ = =

Now, take derivative w.r.t. x 2/x 1,

∴ = b 1/ b 2

ES or θ = *

By reversing b 1/ b 2 :

∴ b 1/ b 2 =

Now substitute this value in the formula of ' θ', i.e. the encircled portion of ' θ' formula = b 2/b 1,

ES = θ = * (dx 2 / dx 1)

But, dx 2 / dx 1 = (b 1 / b 2) . (x 2 / x 1) as mentioned earlier. Again, putting this value inthe function of ES :

ES = θ = *

∴ θ = (b2/ b

1) (x

1 / x

2) (b

1/ b

2) (x

2 / x

1) = 1

d (x 2/x 1) / (x 2/x 1)d (dx 2/dx 1) / (dx 2/dx 1)

d (x 2/x 1)(x 2/x 1)

(dx 2/dx 1)d (dx 2/dx 1)

(dy/dx 1)(dy/dx 2)(b1 y/x 1)(b2 y/x 2)

dx 2dx 1

b1 y x 1

x 2 b2 y

. b 1. x 2 b 2 . x 1

d (dx 2/dx 1)d (x 2 / x 1)

d (x 2/x

1)

(x 2/x 1)(dx

2/dx

1)

d (dx 2/dx 1)

d (x 2/x 1)d (dx 2/dx 1)

(b1/ b 2)(x 2 / x 1)

(b2/ b 1)(x 2 / x 1)

b1 b2

x 2 x 1

.


26/55

Example : Y = 0.7 x 10.3 . x 2

0.7

dy/dx 1 = 0.21 x 1-0.7 . x 2

0.7

dy/dx 2 = 0.49 x 10.3 . x 2

-0.3

∴ = = *

∴ = =

Take derivative of dx 2/dx 1 w.r.t. x 2/x 1 ,

∴ = 3 / 7

Reversing it :

∴ = 7 / 3

Now,ES = θ = * The value of this marked part is 7/3

Putting value 7/3 for marked portion, and also put the value of dx 2/dx 1 = 3x 2/7x 1as given earlier :

∴ ES = θ = *

ES = (7/3) (x 1/x 2) (3x 2/7x 1) = 1 Thus, it is proved that ES is one in C.D. function.

Similarly, one can try for : Y = 2.9x 1

0.5 x 20.7 ,

Y = 1.5x 10.8 x 2

0.2 , Y = 3.2x 1 . x 2

0.5 ,

Y = 0.8x 10.4

. x 2 , Y = 0.5x 1

0.7 x 20.3 ,

Y = 0.8x 10.4 . x 2

0.8 , Y = 0.6x 1

0.80 x 20.6 ,

5. It starts from origin only, which sometime gives wrong estimate, e.g. in case of fertilizer / irrigation experiments, if we do not use fertilizer / irrigation, eventhough

yield will not be zero. Thus, in such cases, this function is not useful.

6. Isoquant are negatively sloped, convex to the origin and asymptotic to the input axes :

(dy/dx 1)

(dy/dx 2)

dx 2dx 1

0.21 x 20.7

x 10.7 x 2

0.3

0.49 x 10.3dx 2dx 1

0.21 x 20.49 x 1

3x 27x 1

d (dx 2/dx 1)d (x 2/x 1)

d (x 2/x 1)d (dx 2/dx 1)

dx 2/dx 1d (dx 2/dx 1)

d (x 2/x 1)(x 2/x 1)

(7/3)(x 2/x 1)

3x 2

7x 1


27/55

Y = ax 1 b1 x 2

b2

x 1 b1 = [y/ax 2

b2 ] ∴ x 1 = [y/ax 2

b2 ] This is the isoquant equation.

Solve this: x 1 = .

Here 'a' is constant and 'y' is also fixed, because isoquant is for given level of output. Thus, the first bracketed part of the right hand side is constant and the coefficient of x 2 isnegative which implies the negative slope.

1/b 1

y a

1 b1 x 2

b2 b1

The isoquant will never touch to the 'x' axisi.e. it is asymptotic to 'x' axis.

Any one isoquant can be derived from theother. It implies that the resources serve aslimitational inputs, i.e. the output becomes zero

when any one of the inputs assumes a zero value, or minimum resources are needed to produce atleast some output.

7. Maxima is not defined :

Y = ax b

, For maxima, equate first derivative with zero,dy/dx = abx b-1 = 0∴ x b-1 = 0/ab = 0∴ x b / x = 0, i.e. x b = 0, x = 0 i.e. not defined.

OR dy/dx = (b . ax b )/ x = (b . y) / x , (ax b = Y)equating this with zero ;∴ dy/dx = (b . y) / x = 0, ∴ by = 0 or Y = 0,i.e. the value of y at any known level of x is not at all defined by this equation.

8. Degrees of freedom (DF) : i.e. n - k - 1, Y = ax 1

b1 x 2 b2 . x 3

b3 ,Here k = 3∴ D.F. = n - 3 - 1 = n - 4

While in quadratic function : Y = a+b 1 x 1 - c 1 x 1

2 + b 2 x 2 - c 2 x 22 + b 3 x 3 - c 3 x 3

2 ± dx 1 x 2 x 3 ,

Here k = 7,∴ Degree of freedom is n - 7 - 1 = n - 8

X 1

X 2OIsoquant


28/55

This shows that the decline in D.F. with the increase in number of variables isrelatively less in Cobb-Douglas function as compared to quadratic function. Hence, Cobb-Douglas production function is very much useful when less number of observations arethere.

9. Slope of the MP is negative :MP = dy/dx = abx b-1 = b . yx -1, Now, the derivative of this will be the slope of MP;

i.e. = - byx - 2 i.e. negative slope.

The negative slope of MPP in C.D. function implies that the MP increases at decliningrate.

10. Marginal Rate of Technical Substitution (MRTS) : Y = ax 1

b1 x 2 b2

∴ dy/dx = ab 1 x 1 b1-1 . x 2

b2 = (b 1 . y) / x 1and dy/dx 2 = ab 2 x 1

b1 . x 2 b2-1 = (b 2 . y) / x 2

∴ = = =

= b 2-1 . b 1 x 2

b2 . x 1 b1-1 . x 1

-b1 . x 2-(b2-1)

= b 1 . b 2-1 . x 1 b1 - 1 - b1 . x 2 b2 - b2 + 1

∴ = ,

Here b 1 and b 2 are fixed, thus, MRTS is a linear function of the ratio (x 2/x 1).

11. Isoclines :By equating MRTS to a constant, say 'k', the isocline equation can be derived as:

= k

∴ x 1 = (1/k) (b 1/ b 2) . x 2

Since is constant, substitute ' ∝ ' for it.

∴ x 1 = ∝ x 2. Thus, the isoclines in C.D. function, is a linear equation withzero intercept. Hence the isoclines are straight lines passing through the origin i.e. Every

isoclines in C.D. function is scale lines. ( ∴ If x 2 = 0, then x 1 = 0). This shows the fixedproportion of mix between the two input variables at different levels of output.

∂2 y ∂ x 2

∆ x 1∆ x 2

(dy/dx 1)(dy/dx 2)

dx 2dx 1

ax 2 b2 . b 1 x 1

b1-1

ax 1 b1 . b 2 x 2

b2-1

dx 2dx 1

b 1 x 1 b 2 x 1

b1 x 2 b2 x 1

b1k b 2


29/55

Isoclines

Isoquants

X 2

X 1

O

12. Ridgelines : When MRTS is equated with zero, it is called ridgelines.

∴ = 0

∴ b 1 x 2 = 0 -- Here b 1 ≠ 0, therefore, if x 2 = 0; then only b 1 x 2 becomes zero. For this reason, the ridgelines in C.D. function will lie only on the axes.

13. Conveniency : This function is practically more convenient since it is linear in log and it

directly gives returnss to scale and elasticity of production. The returns sto scaleindicates the percentage change in output due to one per cent change in all the inputssimultaneously.

14. Maximum profit is obtainable :i.e. maxima is not defined for output, but the maximum profit will be obtained,

MP = dy/dx 1 = ab 1 . x 1 b1 - 1

Now, MR = MC, will give maximum profit and MR = (MP) * price of output i.e. Py similarly, marginal cost = price of input i.e. Px 1,

∴ (MP) . (Py) = Px 1∴ (ab 1 . x 1

b1-1 ) . Py = Px 1

∴ x 1 b1-1 =

∴ x 1 =

* Maxi. profit is possible only when bi


30/55

Similarly,dy dx 1

b 1.y x 1

= = Px 1

Now, taking ratio of both the MPs and equating with price ratio :

= =

∴ b 1 =Px 1 . x 1

y andPx 2 . x 2

y

MPx 1MPx 2

(b1.y/x 1)(b2.y/x 2)

Px 1Px 2

b 1 x 2 b 2 x 1

Px 1Px 2

∴ =

∴ b 1 x 2 = . b 2 x 1Px 1Px 2( (

∴ x 2 = . x 1Px 1

Px 2( ( b2

b1 ((

This is expansion path equation, which is a linear function and starts from origin, hence it is

called scale line.

∴ = . x 2 x 1

Px 1Px 1

b2 b1

This is optimum ratio of the two inputs.

16. Shape of the function :It flattens out as input increases, i.e. a small change in price ratio will shift the

optimum level of input very much, which is ridiculous.

Here, change in theslope of priceline from 'ab' to'cd', has tremendously shiftedthe fertilizer requirement fromOx 1 to Ox 2.

17. Returns to scale : The returns to scale can be easily estimated from the C.D. production function.

Applying this equation in Cobb-Douglas function, Returns to scale = b 1 + b 2 + ................ + b n

= i.e. the summation of the powers of all the input variables (i.e. the summation of all the elasticities)

provides us directly the returns to scale. The returns to scale are decreasing, constant or increasing,

depending on whether ∑ bi is less than, equal to or greater than one.18A. C.D. function allows for only complementary inputs. However, it may be a good approximation for production

processes for which factors are imperfect substitutes over the entire range of X1, X2 values.

n

i = 1∑ bi

Returns to scale = , i = 1, 2, ......, ndy dx i( (

x i y ((∑

b

c d

a

x 1 x 2 Fertilizer

= b 2

Y

O


31/55

18B. C.D. function is a linearly homogeneous function : A production function denoting constant returns to scale is said to be

homogeneous of degree one, indicating that if input is multiplied by constnat amount,the production will be increased in a same ratio. Hence, it is useful to indicate Euler's

Theorem => MP of factor multiplied by the factor quantity for all the factors totalupto the total product.

i.e. y = . x 1 + . x 2dy dx 1

dy dx 1

19. The response function of a Cobb-Douglas production function remains the sameas the original production function since there is no output at x = 0.

Ex.1: Find out fertilizer dose for maximum and optimum output for the followingCobb-Douglas production function :

y = 1.025 x 0.525 Price of fertilizer (Px) = 50 paise per kg.Price of output (Py) = Rs.4 per kg.

y = 10 kg., x = 20 kg.Conditions for maximum output :

(i) = 0 (ii) < 0

Now first we shall try for the maximum output :∴ First condition is = 0

∴ y = 1.025 x 0.525

∴ = (1.025) (0.525) x 0.525 - 1 = 0

= 0.538 x - 0.475 = 0

∴ x = 0

B. Fertilizer dose for optimum production level :Condition for optimum dose is dy/dx = Px/Py,∴ (1.025) * (0.525) - 0.475 = 0.5/4

= 0.538 / x 0.475 = 0.125= 0.538 / 0.125 = x 0.475

∴ x = [0.538 / 0.125] = 21.65 kg.dose of fertilizer for optimum production.

−−

dy dx 1

dy dx 1

dy dx

dy dx

i.e. No maximum output is possible because zero level of input meanszero output, moreover, no secondderivative is possible

Substituting this value of x in the condition of optimality :

i.e. MR = MC OR (MP) (Py) = Px ∴ = {(1.025) (0.525) x - 0.475 } (4) = 0.50∴ = {(0.538) (21.65) - 0.475 } (4) = 0.50 ∴ = 0.4996 ~ 0.50

1/0.475


32/55

Ex.2: Table

Sr.

y

x1

x2

Log x

1

Log x

2

Log y Log x

12

Log x

22

Log y

2 (Log x

1)* (Log x

1)* (Log x

2)*

No.(output(un

it)(unit)

(Log x

2) (Log y) (Log y)

in unit)

1

3.16

2.96

2.29

1.0852

0.8286

1.1506

1.1777

0.6866

1.3239

0.8992

1.2486

0.9533

2

5.19

5.19

2.34

1.6467

0.8502

1.6467

2.7116

0.7228

2.7116

1.4000

2.7116

1.4000

3

9.60

8.92

1.96

2.1883

0.6729

2.2618

4.7887

0.4528

5.1157

1.4725

4.9495

1.5220

4

6.50

3.83

2.14

1.3429

0.7608

1.8718

1.8034

0.5788

3.5036

1.0217

2.5136

1.4241

5

3.68

3.55

0.90

1.2669

-0.1054

1.3029

1.6050

0.0111

1.6975-0.1353

1.6506

-0.1373

6

1.24

3.00

1.68

1.0986

0.5188

0.2151

1.2069

0.2692

0.0463

0.5700

0.2363

0.1116

7

5.44

2.47

1.42

0.9042

0.3507

1.6938

0.8176

0.1230

2.8690

0.3171

1.5315

0.5940

8

6.73

7.43

1.51

2.0055

0.4121

1.9066

4.0200

0.1698

3.6351

0.8265

3.8237

0.7857

9

3.86

2.91

1.02

1.0682

0.0198

1.3507

1.1410

0.0004

1.8244

0.0212

1.4428

0.0267

10

10.33

6.31

2.27

1.8421

0.8198

2.3351

3.3933

0.6721

5.4527

1.5102

4.3015

1.9143

11

5.03

5.01

1.53

1.6114

0.4253

1.6154

2.5966

0.1809

2.6095

0.6853

2.6031

0.6870

12

1.89

3.99

1.05

1.3838

0.0488

0.6366

1.9149

0.0024

0.4052

0.0675

0.8809

0.0310

13

3.60

4.27

1.18

1.4516

0.1655

1.2809

2.1071

0.0274

1.6407

0.2402

1.8594

0.2120

14

6.21

3.66

2.86

1.2975

1.0508

1.8262

1.6835

1.1042

3.3350

1.3634

2.3695

1.9190

15

5.82

4.02

2.57

1.3913

0.9439

1.7613

1.9357

0.8909

3.1022

1.3132

2.4505

1.6625

Total78.28

67.52

26.72

21.5843

7.762522.855432.9031

5.8924

39.272411.5745

34.573113.1059


33/55

∑ log x 12 = ∑ log x 1

2 -

= 32.9031 - ∴ ∑ log x 12 = 1.8443

(∑ log x 1)2

n(21.5843) 2

15

Corrected Values :

∑ log x 22 = ∑ log x 2

2 -

= 5.8924 - ∴ ∑ log x 22 = 1.8753

(∑ log x 2)2

n(7.7625) 2

15

∑ log y 2 = ∑ log y 2 -

= 39.2724 - ∴ ∑ log y 2 = 4.4478

(∑ log y) 2

n

(22.8554)2

15

∑ log x 1 log y = ∑ log x 1 log y --

= 34.5731 - = 1.6852

∑ log x 1 ∑ log y n

(21.5843) (22.8554)15

∑ log x 2 log y = ∑ log x 2 log y --

= 13.1059 - = 1.2782

∑ log x 2 ∑ log y

n(7.7625) (22.8554)

15

∑ log x 1 log x 2 = ∑ log x 1 log x 2 -

= 11.5745 - = 11.5745 - 11.1699

∴ ∑ log x 1 log x

2= 0.4046

∑ log x 1 ∑ log x 2n

(21.5843) (7.7625)15

b1^ =

=

=

= 0.8022

(∑ log x 1 log y . ∑ log x 22) - ( ∑ log x 2 log y . ∑ log x 1 log x 2)

[{∑ log x 12 * ∑ log x 2

2} - (∑ log x 1 log x 2)2 ] = D

{(1.6852) (1.8753)} - {(1.2782) (0.4046)}{(1.8443) * (1.8753)} - {0.4046) 2}

b1^

2.64313.2949


34/55

Similarly,

=

=

= , = 0.5085

b2^ (∑ log x 2 log y . ∑ log x 1

2) - ( ∑ log x 1 log y) . ( ∑ log x 1 log x 2)D

{(1.2782) (1.8443)} - {(1.6852) (0.4046)}

3.2949

b2^1.6756

3.2949

For Standard Errors (S.E.) :∑e2 = ∑ log y 2 - b 1 ∑ log x 1 . log y - b 2 . log x 2 . log y

= (4.4478) - {(0.880) (1.6852)} - {0.6468) (1.2782)}= 4.4478 - 1.4965 - 0.8267

∴ ∑ e2 = 2.1246σ 2 = ∑e2/(n - k - 1) = (2.1246) / (15 - 2 - 1)

∴ σ 2 = 0.1771

Variance of =

= ∴ Variance of = 0.1008

b1^ σ

2 ∑ log x 22

D(0.1771) (1.8443)

3.2949 b1^

Variance of =

=

= ∴ Variance of = 0.0991

b2^ σ

2 ∑ log x 12

D(0.1771) (1.8773)

3.2949

b2^0.3266

3.2949

S.E. ( ) = Var.

= 0.1008

S.E. ( ) = 0.3175

b1^ b1

^

b1^

S.E. ( ) = Var.

= 0.0991

S.E. ( ) = 0.3148

b2^ b2

^

b2^

Cal 't' ratios :

For =

=

= 3.1496

b1^

S.E. of

b1^

b1^

0.80220.3175

For =

=

= 1.6153

b2^

S.E. of

b2^

b2^

0.50850.3184

∴

R 2 = 0.45


35/55

R 2 =

=

= = 0.4500 ∴ R 2 = 0.45 = 45 %

b 1 ∑ log x 1 . log y + b 2 ∑ log x 2 . ∑ log x 2 log y ∑ log y 2

{(0.8022) (1.6852)} + {(0.5058) (1.2782)}4.4478

2.00184.4478

^^

This shows that about 45 per cent variation has been explained by the selectedexplanatory variables.

3. Quadratic function : The word quardratic is originally derived from the Latin word 'Quadratus'

which means squared it.

The function is : Y = a + b 1 x 1 + b 2 x 2 + b 3 x 1

2 + b 4 x 22 + b 5 x 1 x 2 This last term indicates

interaction effect.

Decreasing rate i.e. a + b 1 x 1 - b 3 x 12

Increasing rate i.e. a - b 1 x 1 + b 3 x 12

Characteristics :1) MP is a linear function :

e.g. Y = a + b 1 x 1 ± b 2 x 12 Here the output first increases and then declines,

MP = dy/dx 1 = b 1 ± 2b 2 x 1 = 0, i.e. Linear function.

2) Maxima is defined : Y = a + b 1 x 1 - b 2 x 1

2

∴ dy/dx 1 = b 1 - 2b 2 x 1 = 0∴ b 1 = 2b 2 x 1∴ x 1 = b 1 /2b 2 = 0.5 b 1 b 2

- 1 This is the level of 'x' at total product is maximum.

3) Slope of MP is constant :e.g. MP = dy/dx 1 = b 1 - 2b 2 x 1

∴ slope of MP = - 2b 2 i.e. b 2 is constant,dy 2

∂ x 2

Here, the negative signof the slope of MP impliescontinuous decline of MP or constant decline of MP.

4) Maximum profit :Condition is MP = Px 1/ Py OR

For meaningfulinterpretaion, theconditionsb1, b2>0, and

b3, b4


36/55

(MP) (Py) = Px 1∴ (b1 2b 2 x 1) (Py) = Px 1∴ b 1 . Py 2b 2 x 1 . Py = Px 1∴ b 1 . Py - Px 1 = 2b 2 x 1 . Py

5) Elasticity of production : (EP) :

EP = . =

and MP = dy/dx = b 1 2b 2 x 1 or = b 1 2b 2 x

AP = =

∴ EP = =

±

±

∴ x 1 = b 1Py - Px 1

2b 2Py

This level of x 1 gives us maximum profit.

dy dx

x y

−−

MP AP

± ±

y x

a + b 1 x 1 - b 2 x 12

x 1

MP AP

(b1 - 2b 2 x 1) (x 1)(a + b 1 x 1 - b 2 x 1

2)

Thus, when x 1 increases, EP declines i.e.EP is not constant as it is in case of C.D.

function.

6) Isoquant : Y = a + b 1 x 1 + b 2 x 2 b 3 x 1

2 b 4 x 22 + b 5 x 1 x 2

± ±

Here, for each level of output, there will be two values of input x

1, e.g. A & B, as well

as C & D.e.g. For single variable, the function

may be written as: Ax 2 - Bx - C = 0,

BD

C

A

Where 'C' is constant in respect to X, 'A' is coefficient of X 2 and 'B' is coefficient of X. The following formula is used for computing the value of X :

∴ X = - b ± b2 - 4ac

2a

Similarly, for two variables also it can be calculated as under :

∴ X 1 =- b ± b 2 - 4ac

2a

OR Y - a - b 2 x 2 - b 4 x 22 - b 1 x 1 - b 5 x 1 x 2 - b 3 x 1

2 = 0

This terms is constant i.e. 'C' in relation to x 1

Simplifying it:- (b 1+b5 x 2)x 1

This breaketedpart is coeffi-cient 'b'

This 'b 3 ' is thecoeffi-cient 'a'

b 1 x 1 - 2b 2 x 12

a + b 1 x 1 - b 2 x 12∴ EP =

e.g. Y = a + b 1 x 1 + b 2 x 2 + b 3 x 12 + b 4 x 2

2 + b 5 x 1 x 2


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Now, substituting these values in the algebraic equation :

X 1 =- b ± b 2 - 4ac

2a

∴ X 1 =- (b 1 + b 5 x 2) ± (b 1 + b 5 x 2)

2 - 4b 3 (Y - a - b 2 x 2 - b 4 x 22)

2b 3 This is an isoquant equation. Here, if we put either any of its as zero, eventhough,

some value will be there for x 1 i.e. The factors are independent. Moreover, it is not a linear function. In this way, the value of x 2 can be computed.

N.B.: Here two roots will be obtained, so, minimum value root should be selected.

7) MRTS : Y = a + b 1 x 1 + b 2 x 2 - b 3 x 1

2 - b 4 x 22 + b 5 x 1 x 2

dy

dx 1= b 1 - 2b 3 x 1 + b 5 x 2

dy dx 2

= b 2 - 2b 4 x 2 + b 5 x 1

MRTS = = = =MP of x 1MP of x 2

dy/dx 1dy/dx 2

dx 2dx 1

b 1 - 2b 3 x 1 + b 5 x 2 b 2 - 2b 4 x 2 + b 5 x 1

8) Isoclines :i.e. equating MRTS with some constant value,

MRTS = = k dx 2dx 1

∴ = = k

∴ b 1 - 2b 3 x 1 + b 5 x 2 = k b 2 - 2kb 4 x 2 + b 5kx 1 b 1 + b 5 x 2 - kb 2 - 2kb 4 x 2 = b 5kx 1 + 2b 3 x 1

∴ x 1 (b 5k + 2b 3) = (b 1 - kb 2) + (b 5 + 2kb 4) x 2

dx 2dx 1

b 1 - 2b 3 x 1 + b 5 x 2 b 2 - 2b 4 x 2 + b 5 x 1

∴ x 1 = + . x 2(b1 - kb 2)(b

5k + 2b

3)

(b5 + 2kb 4)(b

5k + 2b

3){ { { {

α β

∴ x 1 = α + β x 2 , Thus,isocline is a linear function, but isoquant does not.

X

Y

Oα

b 1- 2b 3 x 1 + b 5 x 2 b 2 - 2b 4 x 2 + b 5 x 1

∴ x 1 = = + . x 2 b 1 + b 5 x 2

2b 3

9) Ridgelines :i.e. Equate MRTS with zero.

= 0

∴ b 1 - 2b 3 x 1 + b 5 x 2 = 0

b12b 3

b52b 3( ( ( (

∝ β∴ x 1 = ∝ + β x 2 Since both are constant

{


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This shows that the ridgelines are linear function in a quadratic production function.If x 2 = 0, then x 1 = ∝ , i.e. intercept.

The sign of the last term of the production function i.e. b 5 indicates the interaction between x 1 & x 2 ;

If b 5 > 0, => Positive slope of ridgelines, it means positiveinteraction.

If b 5 = 0 => Ridgeline has zero slope i.e. parallel to axes.=> No interaction.

If b 5 < 0 => Negative slope => Negative interaction. => Competitiveness of inputs

Ex.1 : Y = 2500 + 12x - 0.03x 2 WhereY = Output of bajra in kg/ha. x = Fertilizer in kg/ha.

Find out :(i ) AP when x = 50 kg. and x = 140 kg.(ii) MP when x = 80 kg. and x = 120 kg.(iii) EP of x at x = 120.

Ans. (i) AP = Y/x, and Y = 2500 + 12x - 0.03x 2

Now when x = 50, Y = 2500 + 12(50) - 0.03(50) 2

Y = 2500 + 600 - 75 = 3025 kg.∴ AP = Y/x = 3025/50 = 60.50 kg.

When x = 140 kg, Y = 2500 + 12(140) - 0.03(140) 2

∴ Y = 2500 + 1680 - 588 = 3592 kg.∴ AP = Y/x = 3592/140 = 25.66 kg.

Ans. (ii) MP = 12 - 0.06x when x = 80, MP = 12 - 0.06(80) = 7.2 kg.similarly, if x = 120, MP = 12 - 0.06(120) = 4.8 kg.

Ans. (iii) EP = MP/AP , At x = 120, MP = 4.8 kg. AP at x = 120 kg, Y = 2500 + 12(120) - 0.03(120) 2

∴ Y = 2500 + 1440 - 432 = 3508 kg.∴ AP = Y/x = 3508/120 = 29.23 kg.∴ EP = MP/AP = 4.80/29.23 = 0.1642%

x 1

x 2 x 1

x 2

x 1

x 2

Thus, the ridgelines are rectangular in nature and intersect at the maximum output level.10) This function is not homogenous.


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Ex. 2 : The phosphorus response function for groundnut on per hectare basis is : Y = 1800 + 12x - 0.10x 2

(i) Work out the level of x where Y is maximum. What is the maximum attainablelevel of yield ?

(ii) How much quantity of phosphorus should be used if it is free of cost andoutput (y) has some positive price ?

(iii) What are the levels of phosphorus and output at the inflection point ?

Ans. Y = 1800 + 12x - 0.1x 2

(i ) For maximum 'Y', MP = 0∴ dy/dx = 12 - 0.2x = 0∴ 0.2x = 12, ∴ x = 60 kg.

This shows that by using 60 kg of phosphorus, maximum output is achieved.Maximum output (Y) = 1800 + 12 (60) - 0.1 (60) 2

= 1800 + 720 - 360 = 2160 kg.

(ii) If the phosphorus is supplied at free of cost, the aim of the farumers is to get maximum production, (which automatically will give maximum profit too).Hence, to achieve maximum yield of 2160 kg, the quantity of fertilizer neededis 60 kg => The answer is 60 kg of phosphorus should be used even if it isprovided free of cost.

(iii) At the point of inflection, MP is maximum and for that,dy/dx = 12 - 0.2x = 0 and ∂2 y / ∂ x 2 = - 0.2, i.e. < 0, putting this value of - 0.2 in the original equation.

Y = (1800) + 12 (- 0.20) - 0.10 (0.04)∴ Y = 1800 - 2.40 - 0.004 = 1797.596 kg. when MP is maximum.

Allocation of resources among different crops :Problem :

A farmer has got 360 kg. of nitrogen N which he wants to allocate between threecrops Y 1, Y 2 and Y 3 . The area under Y 1 = 3 hectares, Y 2 = 2 hectares and Y 3 = 2 hectares.

Allocate the N among three crops given the following response functions : Y 1 = 15 N 1 - 0.05 N 1

2

Y 2 = 15 N 2 - 0.075 N 22

Y 3 = 15 N 3 - 0.1 N 32

Assume that the price of Y 1, Y 2, Y 3 and N are same i.e. Re 1 / unit.Solution :

First we have to find out whether this is limited resource case or unlimited resourcecase. For this first we will find the marginal productivities by taking partial derivatives

with respect to N 1, N 2 and N 3 and setting equal to zero.


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MPP N 1 = = 15 - 0.1 N 1 = 0 ............... (I)

MPP N 2 = = 20 - 0.15 N 2 = 0 ............... (II)

MPP N 3 = = 25 - 0.2 N 3 = 0 ............... (III)

From equations (i), (ii), (iii) we get 0.1 N 1 = 15 0.15 N 2 = 20

N1 = 150 N 2 = 133.330.2 N 3 = 25 ........> N 3 = 125It is given that 3 N 1 + 2 N 2 + 2 N 3 = 360 ..................... (IV)

We will substitute the values of N 1, N 2 and N 3 in equation (IV) or multiplying by thehectarage under each crop, we will get the total quantity of nitrogen required.

∂ Y 1∂N1∂ Y 2∂N2

∂ Y 3∂N3

3 N 1 + 2 N 2 + 2 N 3= (150 x 3) + (133.33 x 2) + (125 x 2)= 450 + 266.66 + 250= 966.66 kg.

As the total requirement exceeds theinput availability i.e. 966.66 > 360 andhence this is the case of limited resourceavailability.

For allocating the limited resource among three crops, we have to equate the ratioof marginal value product and price of nitrogen in each of the crops.

= =MVPY 1P N 1

MVPY 2P N 2

MVPY 3P N 3

MVP = (MP) . PY

= or MP =

∴

∴

MVPP N 1

MP.PY P N 1

PNPY

As the prices of Y 1, Y 2, Y 3 and N are the same i.e. Re 1 / unit, we can equate the

marginal product of N 1, N 2 and N 3.MPP N 1 = MPP N 2 = MPP N 315 - 0.1 N 1 = 20 - 0.15 N 2 = 25 - 0.2 N 315 - 0.1 N 1 = 20 - 0.15 N 20.1 N 1 = 0.15 N 2 - 5N1 = 1.5 N 2 - 5020 - 0.15 N 2 = 25 - 0.2 N 30.2 N 3 = 5 + 0.15 N 2N

3 = 25 + 0.75 N

2

We have given 3 N 1 + 2 N 2 + 2 N 3 = 360


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Substituting the values of N 1 and N 3 in terms of N 2 we get 3(1.5 N 2 - 50) + 2 N 2 + 2 (25 + 0.75 N 2) = 3604.5 N 2 - 150 + 2 N 2 + 50 + 1.5 N 2 = 360

8 N 2 - 100 = 3608 N

2 = 360 + 100 = 460

N2 = 57.5 kg.N1 = 1.5 N 2 - 50 N 3 = 25 + 0.75 N 2

= 1.5 (57.5) - 50 = 25 + 0.75 (57.5)= 86.25 - 50 = 25 + 43.125

N1 = 36.25 kg. N 3 = 68.125 kg. Total N 1 = 36.25 x 3 = 108.75 kg. " N 2 = 57.5 x 2 = 115.00 kg. " N 3 = 68.125 x 2 = 136.25 kg.

3 N 1 + 2 N 2 + 3 N 3 = 360108.75 + 115.00 + 136.25 = 360 kg.

Thus, the limited quantity of 360 kg. of nitrogen is allocated among 3 crops. Y 1 crop = 108.75 kg. Y 3 crop = 136.25 kg. Y 2 crop = 115.00 kg.

4) Square root function : This function represents a compromise between Cobb-Douglas function and the

quadratic functions. It is sometimes used in response function. The response functionmeans the function is used in deviation form, i.e. without intercept.

Y

O X

Y = f (x 1)

The form is Y = a + b 1 x 1

0.5 + b 2 x 1 for single input Y = a + b 1 x 1 + b 2 x 2 + b 3 x 1

0.5 + b 4 x 20.5 + b 5 x 1

0.5 x 20.5

i, j = 1, 2, ........, n (i < j)

Characteristics :1) Average product (AP) :

AP with respect to x 1 input can be found out as: AP = a + b 1 x 10.5 + b 2 x 1

x 1


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∴ AP = b 2 + a/x 1 + b 1 x 1- 0.5

But in response function i.e. Y = b 1 x 10.5 + b 2 x 1,

AP = = b 2 + b 1 x 1- 0.5 Generally, b 2 < 0 and b 1 > 0

2) Marginal product (MP) :In a single variable function : MP = dy/dx 1 = b 2 + 0.5 b 1 x 1

-0.5

Thus, MP may be very large at low levels of input i.e. x 1, and it declines at slower rate as the input level is increased, e.g. fertilizer => MP is changing over time.e.g. MP = b 2 + 0.5 b 1 x 1

-0.5 if x 1 = 4, b 1 = 0.8, b 2 = 1.9then,MP = (1.9) + (0.5) (0.8) (4) -0.5

= (1.9) + {(0.5) (0.8) / 4 }MP = (1.9) + {0.4 / 2} = 2.10Now, if x 1 = 64,MP = (1.9) + {(0.5) (0.8) (64) -0.5 }

= (1.9) + {(0.5) (0.8) / 64 }MP = (1.9) + {0.4 / 8} = 1.9 + 0.05 = 1.95

This shows that with the increase in the value of X, MP declines.

3) Maxima is defined :

i.e. When MP = 0, TP is maximum.MP = b 2 + 0.5 b 1 x 1

-0.5 = 0∴ 0.5 b 1 x 1

-0.5 = - b 2

∴ x 1-0.5 =

b 1 x 10.5 + b 2 x 1 x 1

- b 20.5 b 1

∴ x 1 = = =

∴ x 1 = = 0.25 b 12 b 2

- 2

- b 20.5 b 1

1-0.5 - b 2

0.5 b 1

0.5 b 1- b 2

-2 +2

0.25 b 1 b1

2

4) Isoquant : Y = a - b 1 x 1 - b 2 x 2 + b 3 x 1

0.5 + b 4 x 20.5 + b 5 x 1

0.5 x 20.5

∴ Y - a + b 1 x 1 + b 2 x 2 - b 3 x 10.5 - b 4 x 2

0.5 - b 5 x 10.5 x 2

0.5 = 0∴ Y - a + b 2 x 2 - b 4 x 2

0.5 - b 3 x 10.5 - b 5 x 1

0.5 x 20.5 + b 1 x 1 = 0

c b a

This is the level of input which yields maximum level of output.

Here, the coefficient of 'X' i.e. b = (- b 3 - b 5 x 20.5 ) X 1

0.5 taking X 10.5 outside.


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∴ x =

∴ x 1 =

Thus, the isoquant equation is not a linear function

5) Elasticity of production (EP) :

EP = . ∴ EP = [b 1 + 0.5b 2 x 1-0.5 ] . x 1/y

Thus, EP is not a constant, but it is a function of x => with the change in value of x, EP will also be changed.

6) R.T.S. : Y = a - b 1 x 1 - b 2 x 2 + b 3 x 1

0.5 + b 4 x 20.5 + b 5 x 1

0.5 x 20.5

MP of x 1 = dy/dx 1 = - b 1 + 0.5b 3 x 1- 0.5 + 0.5b 5 x 1

- 0.5 x 20.5

MP of x 2 = dy/dx 2 = - b 2 + 0.5b 4 x 2- 0.5 + 0.5b 5 x 1

0.5 x 2- 0.5

RTS = =

RTS = i.e. It is not constant.

- b ± b 2 - 4ac2a

+ b 3 + b 5 x 20.5 ± (b 3 + b 5 x 2

0.5 )2 - 4b 1 (Y-a+b 2 x 2 - b 4 x 20.5 )

2b 1

dy dx 1

x 1 y

(dy/dx 1)(dy/dx 2)

dx 2dx 1

- b 1 + 0.5b 3 x 1- 0.5 + 0.5b 5 x 1

- 0.5 x 20.5

- b 2 + 0.5b 4 x 2- 0.5 + 0.5b 5 x 1

0.5 x 2- 0.5

7) Isoclines :Equate RTS with 'k' i.e. constant.

∴ = = k - b 1 + 0.5b 3 x 1

-0.5 + 0.5b 5 x 1-0.5 x 2

0.5

- b 2 + 0.5b 4 x 2-0.5 + 0.5b 5 x 1

0.5 x 2-0.5

Here the isoclines passthrough the origin, it is non-linear and converge to a singlepoint in the input plane.

8) Ridgelines : The R.T.S. is equated with zero.

∴ = 0

∴ - b 1 + 0.5b 3 x 1- 0.5 + 0.5b 5 x 1

- 0.5 x 20.5 = 0

∴ 0.5x 1- 0.5 (b 3 + b 5 x 2

0.5 ) = b 1

∴ 0.5x 1- 0.5 =

∴ x 1- 0.5 =

- b 1 + 0.5b 3 x 1- 0.5 + 0.5b 5 x 1

- 0.5 x 20.5

- b 2 + 0.5b 4 x 2- 0.5 + 0.5b 5 x 1

0.5 x 2- 0.5

b1 b 3 + b 5 x 2

0.5

2b 1 b 3 + b 5 x 2

0.5

x 1 =2b 1

b 3 + b 5 x 20.5

x 1 =2b 1

b 3 + b 5 x 20.5

x 1 = b 3 + b 5 x 20.5

2b 1∴

∴

-2

2

dy/dx 1 = b 1 + 0.5b 2 x 1-0.5

10.5


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Exercise : A] A square root polynomial function is given as below :

Y = 1500 + 60 N 0.5 + 400 P 0.5 - 4 N - 30 P + 7 N 0.5 P 0.5

Where, Y = Yield in kg per hectareN = Nitrogen in kg per hectareP = Phosphorus in kg per hectarePrice of output is Re 1 / unit Price of nitrogen is Rs. 24 per kgPrice of phosphorus is Rs. 35 per kg

Find out the marginal product for N and P and check the second order condition for profit maximization. Find out the optimum levels of N and P and also the ridge line andexpansion path equations.

Answer : Y = 1500 + 60 N 0.5 + 400 P 0.5 - 4 N - 30 P + 7 N 0.5 P 0.5

MP of N = ∂ Y / ∂N = 30 N - 0.5 - 4 + 3.5 N - 0.5 P 0.5

MP of P = ∂ Y / ∂P = 200 P - 0.5 - 30 + 3.5 N 0.5 P -0.5

Profit maximizaation : TR = PY C = r 1 x 1 + r 2 x 2π = TR - C = PY - r 1 N - r 2 P, since x 1 = N and x 2 = P

π = 1 (1500 + 60 N 0.5 + 400 P 0.5 - 4 N - 30 P + 7 N 0.5 P 0.5 ) - 24 N - 35 Pπ = 1500 + 60 N 0.5 + 400 P 0.5 - 4 N - 30 P + 7 N 0.5 P 0.5 - 24 N - 35 P

First order condition for π maximization.∂π / ∂N = 30 N -0.5 - 4 + 3.5 N -0.5 P 0.5 - 24 = 0 .............. (1)∂π / ∂P = 200 P -0.5 - 30 + 3.5 N 0.5 P -0.5 - 35 = 0 .............. (2)

F 1 = r 1, F 2 = r 2

For finding the values of N and Psolve equation (1) and (2)

200P

- 30 + - 35 = 03.5 N

P

30N

- 4 + - 24 = 03.5 P

N

30 - 28 N + 3.5 P = 0 ............. (1)200 - 65 P + 3.5 N = 0 ............. (2)

Multiplying equation 2 by 8 and adding both the equations :

30 - 28 N + 3.5 P = 01600 + 28 N - 520 P = 0

By L.C.M.


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1630 - 516.5 P = 0516.5 P = 1630 P = 1630 / 516.5 = 3.1559 P = 9.95943

Now putting the value of P in equation (1)

we get 30 - 28 N + 3.5 3.1559 = 030 - 28 N + 11.045496 = 028 N = 41.045496 N = 1.465910 N = 2.148893

Second order condition for

maximizationF 11 = - 15 N

- 1.5 - 1.75 N - 1.5 P 0.5

F 22 = - 100 P1.5 - 1.75 N 0.5 P - 0.5

F 21 or F 12 = 1.75 N- 0.5 P - 0.5

By substituting the values of N and P we get F 11 = - 6.51

F 22 = - 3.263F 21 or F 12 = 0.38

Putting the values in B.H. determinant: - 6.51 0.38 - 24By expanding

-6.51 (-1225) - (+0.38) (- 840) - 24 (- 13.3 - 78.24)= 7974.75 + 319.2 + 2196.96= 10171.71

Hence, the second order condition is satisfied.Expansion path :

F 1F 2

r 1r 2

30 N - 0.5 - 4 + 3.5 N - 0.5 P 0.5

200 P - 0.5 - 30 + 3.5 N 0.5 P - 0.5= 24

30

900 N -0.5 - 120 + 105 N -0.5 P 0.5 = 4800 P -0.5 - 720 + 84 N 0.5 P -0.5 = 0900 N -0.5 - 4800 P -0.5 + 105 N - 0.5 P 0.5 - 84 N 0.5 P -0.5 + 600 = 0

This is the equation of expansion path.Ridgeline :

PN

30 N - 0.5 - 4 + 3.5 N - 0.5 P 0.5

200 P - 0.5 - 30 + 3.5 N 0.5 P - 0.5= = 0

= 30 N -0.5 - 4 + 3.5 N -0.5 P 0.5 = 0= 200 P -0.5 - 30 + 3.5 N 0.5 P - 0.5 = 0

5) Mitscherlich - Spillman Production function :

E.A. Mitscherlich is the first Economist to explore the nature of fertilizer responsefunction in 1909. The suggested equation is : log ao - log (ao - Y) = CX

NP


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Where 'ao' is the maximum attainable yield with the addition of amount of nutrient;'C' is proportionality constant, which defines the rate at which marginal yields decline and

Y is the yield response. This function was severely criticised since it did not allow negative marginal

products. Moreover, to think of 'C' as constant irrespective of crop is also net reasonable.

6) Spillman function : Y = µ - AR x , Where µis the maximum attainable yield by increasing the level of nutrient, x; A is

a constant which represents the maximum response attainable from the use of input x; andR = MP i / MP i-1;

The above function can also be written as : Y = M - A + A - AR X = (M-A) + A (1 - R X ) {Adding & subtracting 'A'} Where (M - A) is the output obtained without any use of the input, and A (1 - R X ) is

the addition to the yield or response. The function was further developed as under : Y = A (1 - R 1

X1) ; Y = A (1 - R 1

X1) (1 - R 2 X2) and y = A (1 - R i

Xi)n

i = 1π

Y

O X

Y = M - AR X

M - A

M

Spillman function

Characteristics :1) In pu t o utp ut cur ve isasymptotic,

2) AP x1 =

= A (1- R 1 x1 ) x 1

- 1

It means AP of x 1 is a functionof its own level.

A (1-R 1 x1

) x 1

3) MP curve is asymptotic to the input axis and never assuming a negative value.

MP of x 1 = dy/dx 1 = - AR x1 . log R 1, Thus, this function cannot be used in the situations where there is over utilisation

of resources.

4) Isoquants : Y = A (1 - AR 1

x1 ) (1 - AR 1 x2 )

x 1 = log 1 - (log R 1)- 1

This implies that one input can never substitute completely for the other.

Y A (1 - R 2

x2 ) {{


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5) R.T.S. :

RTS of x 2 for x 1 = =dx 1dx 2

(1- R 1 x1 ) R 2

x2 . log R 2(1- R 2

x2 ) R 1 x1 . log R 1

RTS 21 is a function of the levels of x 1 and x 2 inputs.

6) Isoclines :By equating RTS equation to constant 'k' ;

= k

∴ = k (1- R 2 x2 ) R 1

x1 . log R 1 = (1- R 1 x1 ) R 2

x2 . log R 2= R 2

x2 . log R 2 - R 1 x1 . R 2

x2 . log R 2∴ k (1- R 2 x2 ) R 1 x1 . log R 1 + R 1 x1 . R 2 x2 . log R 2 = R 2 x2 . log R 2

∴ R 1 x1 =

Take natural log on both the sides :

∴ x 1 . log R 1 = log

∴ x 1 = log log R 1

(1- R 1 x1 ) R 2

x2 . log R 2(1- R 2

x2 ) R 1 x1 . log R 1

R 2 x2 . log R 2

k (1- R 2 x2 ) log R 1 + R 2

x2 . log R 2

R 2 x2 . log R 2

k (1- R 2 x2 ) log R 1 + R 2

x2 . log R 2

R 2 x2 . log R 2k (1- R 2

x2 ) log R 1 + R 2 x2 . log R 2

These isoclines are therfore, not straightlines. They pass through theorigin. The slope of the isoclines isasymptotic.

Isoclines

Isoquants

Y

O X

7) Ridgelines :By equating the RTS equation with zero and solving for x 1 in terms of x 2 or vice

versa, it can be seen that the ridgelines are identical with the input axes.

7) Balmukand or Resistance function :B. Balmukand is the first Indian to give the production function. The form of the

function is : Y - 1 = a (b + x) - 1 + c,

Where a, b & c are parameters. In case of fertilizer fn'b' = nutrient in soil & x = amount added


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49/55

Thus, CES function is a homogeneous of degree one ( power of 't' is one) i.e. it islinearly homogeneous => output increases as same percentage as inputs increase. Thisalso implies that Euler's theorem is qualified here. In case of C.D. fucntion, it ishomogeneous, but it is homogeneous of degree one only when ∑ b i = 1.

∴

e.g. Y = ax 1 b1

. x 2 b2

Now increase x i by 't' times,∴ Y = a (x 1t)

b1 . (x 2t) b2

= a x 1 b1

. t b1 . x 2

b2 . t

b2

= t b1 + b2 . ax 1 b1 x 2

b2

= ( t b1 + b2 ) . Y

Here, b 1 + b 2 = degree of homogeneity. Thus, if b 1 + b 2 = 1, then only the C.D. function becomes homogeneousof degree one.* Quadratic function is not homogeneous.

2) Constant Elasticity of Substitution ( ES or σ ) :Let us assume the two variables 'k' & 'L',

∴ ES = σ = ÷

ES = *

d (K/L)(K/L)

d (dk/dL)(dk/dL)

d (k/L)(k/L)

(dk/dL)d (dk/dL)

Now, the CES function is : y - θ = µ k - θ + ∝ L -θ ,For working out dk/dL, differentiate this function. Here Y = fixed, hence its

derivative is zero.∴ 0 = - θ µ k -θ−1. dk - θ ∝ L -θ−1 . dL ∴ θ µ k -(θ+1) . dk = - θ ∝ L -(θ+1) . dL

Solving this :

∴ =dk dL

θ ∝ L -(1+ θ)

θ µ k -(1+ θ)

∴ MRTS = =dk dL

∝

µL k ( ( ( (

- (1 + θ)

OR ∴ MRTS = = - .dk dL

∝

µK L

1 + θ

( ( ( (

This is same as in C.D.function, except (1 + θ) in power.

Now differentiate the MRTS.i.e. dk/dL w.r.t. k/L,

∴ ∆ in MRTS = ∴ ∆ in MRTS = - (1 + θ)d (dk/dL)

d (k/L)∝

µK L

θ

( (Reversing it, we get ;

= - .d (k/L)d (dk/dL)

µα

θ( (11 + θ L k


50/55


51/55

(

Similarly,dy/dL = - A/ θ [δ k -θ + (1 - δ) L -θ ]- (1+(1/ θ)) * [−θ(1−δ) L -(1+θ)]But δ > 0 and θ ≥ -1, thus, MP is always positive.

5) MRTS :

- But, α = δ (1−α) and µ = δ . α

∴ MRTS = -

= - {(1- α) / α } {k/L}1+ θ

∝

µk L

1 + θ

( (δ (1−α)

δ . αk L

1 + θ

( ({ { This is the Slope of theisoquant for a given factor ratio.Here, MP is having positive values.

Case - I : Assume that θ = -1

∴ dk/dL = - {(1- α ) / α } {k/L}1-1 = {- (1- α ) / α }

i.e. MRTS is constant at eachpoint, because α is constant, or slope of an isoquant is a linear.

O

k

L

Case - II : Suppose θ = α i.e. infinite, Again, here few situations are there :(a) e.g. if θ = α and K/L > 1,

∴ = - = - αdk dL

k L

1 + θ1 − δδ

i.e. infinite, because θ = α, and θis power to k/L and again k/L isgreater than one. L

O

K

(k/L) > 1

(b) If θ = α and k/L < 1, then

= -

e.g. k = 100, and L = 200, i.e. k/L = 1/2,

= -

Here, θ = α ∴ dk/dL => 0, i.e. it tends to zero, or nearer to zero.

dk dL

k L

1 + θ1 − δδ

12

1 + α1 − δδ ( ( (


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(c) If (k/L) = 1 and θ = α ,

∴ = -

∴ = - i.e. constant.

(dk dL 1 + θ1 − δ

δ ( ( (1dk

dL (1 − δ

δ (Case - III : e.g. θ = 0,

= -

= -

∴ = -

(dk dL 1 + θ1 − δ

δ ( ( (k L (

11 − δδ ( ( (k L

dk

dL (1 − δ

δ ( ( (k

L

Thus, if θ = 0, MRTS of CESfunction is just equal to MRTS of C.D.

function [i.e. (b 1/b 2) . (x 2/x 1)].

dk

dL

1

θ( (

6) Elasticity of production :EP for x 1 or k = (dy/dx 1) . (x 1/y)Function is :

Y = A [ δ k - θ + (1 - δ) L -θ] -1/ θ

MP of k = dy/dk

∴ = A - [ t ] - (1 / θ)-1 * [ (- θ) δk - (1+ θ) ]

Where t = δk - θ + (1 - δ) L - θ

∴ = δ A [ t ] - (1 / θ)-1 * k - (θ − 1)

∴ = δ A [ t ] - (1+ θ)/ θ * k - (θ + 1)

∴ = δ [ t ] - [(1+ θ)/ θ] * k - (1+θ)

∴ = { A [ t ] - 1/ θ}1+ θ * k - (1+θ)

Since (- 1/ θ) (1 + θ) = (- 1 / θ) - 1.But value of 't' = δk -θ +(1- δ) L −θ ,substituting this value, A (t) 1/ θ becomes 'Y', i.e. original function.

Thus,= [Y]1+ θ * k - (1+ θ)

∴ =

dy dk dy dk

A 1+ θ

A θ

δ A θ

δ A θ

dy dk

δ A θ

y K

1+θ


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EP of k = *

= *

∴ EP of k =

Similarly, for second input ....

7) L ' Hopital's rule : As per this rule, we derive the function as below : Y −θ = µk −θ + αL −θ , if θ = 0, then Y 0 = µk −0 + αL −0 ,

∴ 1 = µ +α , Now, total differentiation as per this rule for Y −θ

= µk −θ

+ αL −θ

,∴ - θ Y −(1+θ) . dy = - θµk −(1+θ) . dk - θα L −(1+θ) . dL

= − θ = − θ − θα

∴ = µ . + α

dy dk

k Y

δ A θ

Y k

1+θ k Y

δ A θ

Y k

θ

dy Y (1+ θ)

µ.dk k 1+θ

dL L (1+ θ)

dy Y (1+ θ)

dk k 1+θ

dL L (1+ θ)

Now, as per this rule, putting θ = 0.

∴ = µ . + α Integrate this function.dy Y

dk k

dL L

= µ . + αdy Y

dk k

dL L ∫ ∫ ∫

∴ Log Y = µ . log k + α log L + log A i.e. constant,

So, this function is reduced to C.D. function, i.e. Y = A k µ . L α .Now, if θ = -1,∴ Y − θ = µk − θ + α L − θ

Put θ = -1, Y = µk + αL i.e. if θ = -1, then CES becomes linear function. Thus CES function

is based on the value of θ.

9) VES function :i.e. variable elasticity of substitution. This function was given by Ravenker.

θ = b A a (1- dp) [ N + (P - 1) A ] adp

Where A = area, N = labour, θ = output


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Here b > 0 => gives efficiency parameter,a > 0 gives scale parameter i.e. just like µ in CES function.P is the slope.0 ≤ dp ≤ 1 ; ≥

N A

1 - P1 - dp( (

If "a" is removed from the function, then it will be linearly homogeneousproduction function.


55/55

Q.1. Answer the following questions in brief.1) Explain the meaning of fabrication coefficient.2) What do you understand by linearly homogeneous production function ?3) All ridgelines are isoclines but all isoclines are not ridgelines, comment on this

statement.

4) Prove that isoquant is a linear function in case of linear production function.5) Derive the average product, when the square root function is given in deviation

form.6) Find out the optimum level of input for profit maximisation in case of quadratic

function.7) Differentiate the risk and uncertainty.8) What do you understand by saddle point in Game Theory ?9) The Cobb-Doglas production function has greater utility when there is less number

of observations. How ?10) Calculate the elasticity of substitution for the following function :

Y = 1.5 x 10.8 . x 2

0.8

11) What do you understand by capital rationing ? What is the utility of capitalrationing ?

12) Find out the level of input, where total product is maximum in case of square-root function.

13) Explain the variance criteria for diverssification.

Q.2. What is production function ? Explain various assumptions for estimatingproduction function.

Q.3. A] Explain the following characteristics of Cobb-Doglas production function:i) Ridgelines ii) Isoquant iii) Expansion pathiv) Elassticity of production

B] Under which different situations, the Cobb-Doglas production function isapplied ?

Q.4. Discuss in details the returns to scale alongwith its test of significance and suitablediagrams.

Q.5. Derive the equations of marginal rate of technical substitution and ridgelines incase of quadratic function. Also give your comments.

Q.6. Give the graphical presentation of the following :i) Isoclines ii) Isoquant map iii) Ridgelinesiv) Scale line

prod func part - 1

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