Bi-Variate Functions - ACTIVITES

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Bi-Variate Functions - ACTIVITES LO1. Students to consolidate basic meaning of bi-variate functions LO2. Students to learn how to confidently use bi-variate functions in economics Students are given the bivariate function : Y = X α β 1 X 2 They are put into small groups and asked to produce lists of all meaningful economic relationships connecting the variables utility, income, tax rate, consumption, disposable income, output, labour, capital, government revenue, interest rate, investment, apples, oranges, etc. which can be described by the function: A complete answer involves specifying what the labels Y, X 1, and X 2 mean and any restrictions on α and β are. Task 1 Construct one example where α and β are both positive Task 2 Construct one example where α is positive but β is negative Task 3 Construct one example where α and β both take on specific numerical values. LO1. Students to consolidate basic meaning of bi-variate functions LO2. Students to learn how to confidently use bi-variate functions Page 1 of 15

Students are given the bi-variate function : Y = X 2 1 X 3 2 and asked to complete the final column and comment on their ansers. X 1 X 2 Y (i) 5 6? (ii) 5 7? (iii) 3 4? (iv) 3 5? Page 2 of 15

Bi-Variate Functions - ANSWERS There are a large number of solutions and, of course, the purpose is not to create a particular solution but rather to practise the confident use of bi-variate functions as used in economics. Possible solutions include Task 1: Y=utility, X 1 = oranges, X 2 =apples Task 2: Y= Consumption, X 1 = income, X 2 =interest rate Task 3: Y= Government revenue, X 1 = income, X 2 =tax rate, α and β =1 5400 8575 576 1125 Comparing (i) with (ii), note that Y increases by (8575 5400) = 3175 due to a unit increase in (X 2 ) whereas comparing (iii) with (iv) Y increases by only (1125-576)= 549 due to a unit increase in (X 2 ) Message: The absolute change in Y, when only one independent variable changes, depends on the starting level of both independent variables. Page 3 of 15

Partial Differentiation - ACTIVITIES LO1. Students to consolidate basic meaning of partial derivative functions LO2. Students to learn how to confidently calculate partial derivative functions in economics Students are given the bivariate function : Y = X 1 2 X 2 3 which reduces to the univariate function Y=8 X 1 2 when X 2 = 2. 2 Students are divided into two groups. Students in Group A find the ordinary derivative of Y=8 X 1 ( answer is 16 X 1 ) and evaluate it for X 1 = 2, 3, 4 etc. Students in Group B find H 1, the partial derivative of Y= X 2 3 1 X 2 w.r.t. X 1 ( answer is 2X 1 X 3 2 ) and evaluate it for (X 1,X 2 ) = (2,2), (3,2), (4,2) etc They then compare notes by reporting their answers and filling up the following table: Value of X 1 Group A Ordinary derivative of Y=8 Group B H 1,Partial derivative of Y= X 2 3 1 X 2 2 3 4 X 1 2 w.r.t. X 1 (evaluated when X 2 =2) The numerical results in Column 2 and 3 should be the same. Task 1 Create a table similar to the above when X 1 is constant at 3 but X 2 is varying. Task 2 Draw a rough sketch of H 1 against X 1 holding X 2 constant at X 2 =2 Task 3 Draw a rough sketch of H 1 against X 2 holding X 1 constant at X 1 =4 Page 4 of 15

LO1. Students to consolidate the idea that the partial derivative (like the ordinary derivative) is a function (not a number) Students are given the bi-variate function: Y = H(X 1, X 2 ) = X 2 1 X 3 2 and asked to complete the following table and comment on their answers X 1 X 2 H 1 H 2 (i) 5 6?? (ii) 5 7?? (iii) 3 4?? (iv) 3 5?? Page 5 of 15

Partial Differentiation - ANSWERS In the tables, columns 2 and 3 should produce the same answers. Students should appreciate why they are equivalent. Group A is reducing the bi-variate function to a univariate function first and then differentiating whilst Group B is differentiating first and then setting a specific value for the constant X 2. Task Two 100 80 60 40 20 H1 X1 (X2=2) 0-5 -4-3 -2-1 0 1 2 3 4 5-20 -40-60 -80-100 Page 6 of 15

Task Three 1500 1000 500 H1 0-5 -4-3 -2-1 0 1 2 3 4 5-500 -1000-1500 X2 (X1=4) 2160, 2700 3430, 3234 576, 432 750, 675 Comparing (i) with (ii) and (iii) with (iv) note that both H 1 and H 2 numerically increase due to a unit increase in (X 2 ) Message: The value of both partial derivatives (as only one independent variable changes) depends on the starting level of both independent variables. Page 7 of 15

Unconstrained Optimisation - ACTIVITIES LO1. Students to consolidate writing down an objective function which is to be maximised. LO2. Students to learn how to obtain the FOC LO3. Students learn how to solve the FOC Students are given the following problem. A multiproduct firm has cost function given by TC=ax 2 +bxy +cy 2 + K where x and y are the quantities produced of its two outputs. The prices of these are fixed in world markets at p and q respectively. How much of each output should the firm produce? Students should now be split into four groups and each group be given the following tasks. The groups should not be in contact, i.e. each group works in isolation from all others. Group A: Write down the optimisation problem in standard form, making the notation clear. Group B: Given the function π = px + qy [ax 2 +bxy +cy 2 + K], find the partial derivative functions π x and π y. Group C: Given the problem and the partial derivatives π x = p 2ax by and π y = q bx 2cy What are the FOC? Group D: Find the solution to the equations p 2ax by = 0 and q bx 2cy = 0 LO1. Students learn how to operate SOC to check for a maximum : Continuing with the above problem students are now divided again into three groups. Each group works in isolation as before and is assigned the following tasks. Page 8 of 15

Group A: Calculate the second partial derivative functions: π xx, π xy and π yy, π yx Group B: Given the answers to the second partial calculations, find the value of and π xx Group C: Using economic intuition and the fact that = 4ac b 2 and π xx = -2a, discuss whether the SOC are satisfied. Page 9 of 15

Unconstrained Optimisation - ANSWERS Group A : Choose x and y to Maximise π = px + qy [ax 2 +bxy +cy 2 + K] Group B: π x = p 2ax by and π y = q bx 2cy Group C: p 2ax by = 0 and q bx 2cy = 0 Group D: x* = [2cp-bq]/[4ac b 2 ] and y*= [2aq bp]/[ 4ac b 2 ] Then the sequence of answers is put together by the Lecturer and the whole group discusses the question Are the values of x* and y* found by this procedure the firm s optimal output levels? [Answer: We cannot be sure because the Second Order Conditions have not been checked]. Group A: -2a, -b, -2c, - b. Note that π xy = π yx, both being equal to b. Group B: = 4ac b 2 and π xx = -2a Group C : It is immediately obvious that part of the SOC is satisfied because π xx is clearly negative. But what about? Economic intuition tells us that if the price of y, i.e. q goes up, it would be utterly perverse for the firm to sell less y. Similarly for x. In other words, we expect supply curves to be upward sloping! From the solutions to FOC we note that x* = [2cp-bq]/[4ac b 2 ] and y*= [2aq bp]/[ 4ac b 2 ]. The impact of p on x* and of q on y* will be positive only if 2c/[4ac b 2 ] and 2a/[ 4ac b 2 ] is strictly positive. But 2c and 2a are clearly positive. Therefore in order for upward sloping supply curves, we require [4ac b 2 ] to be strictly positive. But [4ac b 2 ] is precisely the value of. Hence, > 0 and the SOC are satisfied. Page 10 of 15

Constrained Optimisation - ACTIVITIES LO1: Students to learn how to obtain the FOC LO2: Students learn how to solve the FOC LO3: Students learn how to interpret the FOC Students are given the following problem: (a) Maximise H= p 1 X 1 + p 2 X 2 s.t. G= X 1 2 + X 2 2 r 2 and (X 1,X 2 ) 0 (b)using your answer to (a) write down the FOC and interpret (c) Using your answer to (c) solve these three equations LO1. Students consolidate solving and interpretation of FOC. This is an individual activity. All students are given the following problem: A small country is seeking to maximise its export revenue by selling its coffee and coconuts on the world market where they fetch $ 6 and $ 8 per ton respectively. In producing coffee and coconuts, the country is constrained by its production set which is given by: X 1 2 + X 2 2 100 Where X 1 denotes coffee production in tons and X 2 denotes coconut production in tons. Find the optimum production values X * 1 and X * 2 of coffee and coconuts and the export revenue earned. At what rate would the export revenue increase if the production set constraint was slightly relaxed. Page 11 of 15

Task Two Re- solve the problem if the production constraint was changed to X 2 1 + X 2 2 121. By how much does export revenue change? How could you obtain this answer without resolving the whole problem? If the change in the production constraint to be shifted to X 2 1 + X 2 2 121 is achievable by buying a technology, what is the most the country should pay for this technology?. ACTIVITY THREE LO1. Students learn how to operate convexity conditions to check for a maximum Students are asked to draw shapes of convex and non convex sets Task Two Students are asked to note common economic problems in which the relevant constraints do indeed form a convex set. Task Three Students are asked to note how to identify convex superior sets that arise commonly in economics. Task Four Students should check whether the convexity conditions are satisfied for the problem above. Page 12 of 15

Constrained Optimisation - ANSWERS 2 (a) Answer: L = p 1 X 1 + p 2 X 2 - λ [ X 1 + X 2 2 - r 2 ] It is very important that students write the Lagrangean correctly, particularly noting the NEGATIVE sign between the objective and the constraint. They also need to note that in this problem r 2 is c. (b) Answer: L 1 = p 1-2 λ X 1 =0 (1) L 2 = p 2-2 λ X 2 =0 (2) 2 λ [ X 1 + X 2 2 - r 2 ]=0 (c) Answer: To solve these three equations we start with 1) and 2). Rewrite (1) and (2) as: p 1 = 2 λ X 1 (4) p 2 = 2 λ X 2 (5) Divide 4) by 5) to get: p 1/ p 2 = X 1 / X 2 (6) From 4) and 5) since both p 1 and p 2 are strictly positive, it follows that (X 1 *,X 2 * ) and λ* are strictly positive. Since λ*>0, it follows that: X 1 2 + X 2 2 - r 2 =0. (7) Now solve 6 and 7 to get: Page 13 of 15

X 1 * = r p 1 / [p 1 2 + p 2 2 ] and X 2 * = r p 2 / [p 1 2 + p 2 2 ] To find the value of λ*, substitute X 1 * = r p 1 2 /[p 1 2 + p 2 2 ] in 4) to get: λ* = { [p 1 2 + p 2 2 ]}/2r Thus the full solution is: X 1 * = r p 1 / [p 1 2 + p 2 2 ] and X 2 * = r p 2 / [p 1 2 + p 2 2 ] and λ* = { [p 1 2 + p 2 2 ]}/2r As a final step substitute (X 1 *,X 2 * ) from above into H to obtain: H* = r p 1 2 / [p 1 2 + p 2 2 ] + r p 2 2 / [p 1 2 + p 2 2 ] TASK ONE This is just a numerical version of the problem analysed in Activity 1. L=6X 1 + 8X 2 λ [X 1 2 + X 2 2 100] FOC are: (1): L 1 = 6-2 λ X 1 = 0 (2): L 2 = 8-2 λ X 2 = 0 (3): λ[x 1 2 + X 2 2 100] = 0 From (1) and (2), ¾= X 1 / X 2 (A) From (A) both X 1 and X 2 are strictly positive. Hence from (1) so is λ >0 hence from (3) X 1 2 + X 2 2 100 = 0 (B) Substitute (A) into (B) to get: (X 1 * = 6, X 2 * = 8 ) and R* = 100. Substitute X 1 * = 6 into (1) to get λ* = ½. Page 14 of 15

The rate at which the export revenue increase if the production set constraint was slightly relaxed = λ* = ½. R* = 100 TASK TWO If the constraint changes to X 1 2 + X 2 2 121, then one can re-solve the whole problem to get: (X 1 ** = 6.6, X 2 ** = 8.8 ) and R ** = 110. Hence dr* = 110-100= 10. Much easier to solve by using the information contained in the Lagrange Multiplier, i.e. dr*/d(r 2 )= λ* Since λ* = ½, and d(r 2 )=121-100=21, then dr *= ½ X 21 = 10.5 which is approximately same as by long method. ACTIVITY THREE Convex sets are triangles, circles, semi circles, quarter circles, rectangles squares etc. The best example of a non-convex set is a doughnut. Task Two Consumer theory ( budget set is a triangle, hence convex), Producer theory ( production function with non-increasing returns to scale) Task Three Indifference curve shapes or iso-profit lines which are linear. Task Four Yes they are since H= p 1 X 1 + p 2 X 2 is linear and hence the Superior set S [H(, ) h ] for any h is a convex set. Furthermore since the Feasible set F is a quarter circle and hence convex. Page 15 of 15