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A1. Milk or meat production Y per day is best explained by the inputs: feed needed per day in kg, X₁, and (number of) milk cows or calves, X2. Based on fitting the data for milk or meat production the revenue is modelled by a Cobb-Douglas function R(X₁, X2) = pX₁ X₂ constrained by relation h(X1, X2) = C₁X1+C2X2 = C3. For milk production, the constraint expresses a relation between the average sale price c3 of milk (per day), the average price c₁ of a kg of feed and the price c₂ of a milk cow (cost per day and based on the annual cost of a milk cow). For meat production, the constraint expresses a relation between the average sale price c3 of meat (per day), the average price c₁ of a kg of feed and the price c₂ of a calf (cost per day and based on the annual cost of a calf). The parameters for the two cases have been established by linear regression of data to be the following: £4.00, C2 = £1.36/d, c3 = £5.38; regarding the milk case: p = 445.69, a = 0.346, b = 0.542, c₁ = and, regarding the meat case: p= 2.348, a = -0.205, b = 1.118, c₁ = £4.00, c₂ = £500.0, c3 = £6.02.

A1. Milk or meat production Y per day is best explained by the inputs: feed needed per day in kg, X₁, and (number of) milk cows or calves, X2. Based on fitting the data for milk or meat production the revenue is modelled by a Cobb-Douglas function R(X₁, X2) = pX₁ X₂ constrained by relation h(X1, X2) = C₁X1+C2X2 = C3. For milk production, the constraint expresses a relation between the average sale price c3 of milk (per day), the average price c₁ of a kg of feed and the price c₂ of a milk cow (cost per day and based on the annual cost of a milk cow). For meat production, the constraint expresses a relation between the average sale price c3 of meat (per day), the average price c₁ of a kg of feed and the price c₂ of a calf (cost per day and based on the annual cost of a calf). The parameters for the two cases have been established by linear regression of data to be the following: £4.00, C2 = £1.36/d, c3 = £5.38; regarding the milk case: p = 445.69, a = 0.346, b = 0.542, c₁ = and, regarding the meat case: p= 2.348, a = -0.205, b = 1.118, c₁ = £4.00, c₂ = £500.0, c3 = £6.02.

ENGR.ECONOMIC ANALYSIS
ENGR.ECONOMIC ANALYSIS
14th Edition
ISBN:9780190931919
Author:NEWNAN
Publisher:NEWNAN
Chapter1: Making Economics Decisions
Section: Chapter Questions
Problem 1QTC
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= A1. Milk or meat production Y per day is best explained by the inputs: feed needed per day in kg, X₁, and (number of) milk cows or calves, X2. Based on fitting the data for milk or meat production the revenue is modelled by a Cobb-Douglas function R(X1, X2) = pX1 X₂ constrained by relation h(X1, X2) C₁X1+C2X2 = c3. For milk production, the constraint expresses a relation between the average sale price c3 of milk (per day), the average price c₁ of a kg of feed and the price c2 of a milk cow (cost per day and based on the annual cost of a milk cow). For meat production, the constraint expresses a relation between the average sale price c3 of meat (per day), the average price c₁ of a kg of feed and the price c₂ of a calf (cost per day and based on the annual cost of a calf). The parameters for the two cases have been established by linear regression of data to be the following: £1.36/d, c3 = £5.38; regarding the milk case: p = 445.69, a = 0.346, b = 0.542, c₁ = £4.00, c₂ and, regarding the meat case: p = 2.348, a = -0.205, b = 1.118, c₁ = £4.00, c2 = £500.0, c3 = £6.02. Consider the general case first (without substituting any parameter values). Determine the Lagrangian. Find the FOCs, stationary point and analyse the bordered Hessian to classify the critical point. (Hint: Do not eliminate either X₁ or X2 till you need to.) For the general case, verify your findings by repeating the analysis by first eliminating X2. Make a comparison of the two analyses. Subsequently use these outcomes to analyse the cases for milk and meat production and draw conclusions. In particular, what improvements in the production and/or mathematical analysis could or should be made, if any?
Transcribed Image Text:= A1. Milk or meat production Y per day is best explained by the inputs: feed needed per day in kg, X₁, and (number of) milk cows or calves, X2. Based on fitting the data for milk or meat production the revenue is modelled by a Cobb-Douglas function R(X1, X2) = pX1 X₂ constrained by relation h(X1, X2) C₁X1+C2X2 = c3. For milk production, the constraint expresses a relation between the average sale price c3 of milk (per day), the average price c₁ of a kg of feed and the price c2 of a milk cow (cost per day and based on the annual cost of a milk cow). For meat production, the constraint expresses a relation between the average sale price c3 of meat (per day), the average price c₁ of a kg of feed and the price c₂ of a calf (cost per day and based on the annual cost of a calf). The parameters for the two cases have been established by linear regression of data to be the following: £1.36/d, c3 = £5.38; regarding the milk case: p = 445.69, a = 0.346, b = 0.542, c₁ = £4.00, c₂ and, regarding the meat case: p = 2.348, a = -0.205, b = 1.118, c₁ = £4.00, c2 = £500.0, c3 = £6.02. Consider the general case first (without substituting any parameter values). Determine the Lagrangian. Find the FOCs, stationary point and analyse the bordered Hessian to classify the critical point. (Hint: Do not eliminate either X₁ or X2 till you need to.) For the general case, verify your findings by repeating the analysis by first eliminating X2. Make a comparison of the two analyses. Subsequently use these outcomes to analyse the cases for milk and meat production and draw conclusions. In particular, what improvements in the production and/or mathematical analysis could or should be made, if any?
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