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: regarding the milk case: p = 445.69, a = 0.346, b = 0.542, C1 = £4.00, c₂= £1.36/d, c3 = £5.38; 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) = 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: regarding the milk case: p = 445.69, a = 0.346, b = 0.542, C1 = £4.00, c₂= £1.36/d, c3 = £5.38; 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.)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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) = pX₁X₂ constrained
by relation h(X1₁, X₂) = C1₁X₁ + 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
and,
£500.0, C3
regarding the meat case: p= 2.348, a = -0.205, b = 1.118, c₁
=
=
£4.00, C₂
£4.00, C₂
=
=
= £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.)
(ii) For the general case, verify your findings by repeating the analysis by first eliminating X₂.
Make a comparison of the two analyses.
(iii) 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?
(iv) Make a sketch/graph of the solution (either made by hand or by using the computer, e.g., by
using Python) for both the milk and meat cases.
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