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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) = pXX 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.) (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.

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) = pXX 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.) (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.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 91E
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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) = 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.
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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