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) = C₁ X₁ + 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: C1 £4.00, C₂ = 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. (i) 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.)
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) = C₁ X₁ + 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: C1 £4.00, C₂ = 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. (i) 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.)
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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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) = C₁ X₁ + 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:
C1
£4.00, C₂ =
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.
(i) 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.)
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