F Q RESEARCH METHODS

TOPIC 2 OPTIMISATION
1. Multi Variables Optimisation 2. Constrained Optimisation I 3. Constrained Optimisation II
Reading: Jacques Ian (2006), Mathematics for Economics and Business, 5th Edition Prentice Hall - Chap 5.1, 5.4, 5.5, 5.6 & Appendix 3 Chiang, A and Wainwright, K (2005), Fundamental methods of mathematical economics, Mc Graw Hill - Chap 12.1, 12.2, 12.3

1. Multi Variables Optimisation
Refresher: Optimisation of a function
 Suppose have a function like this: f(x) = 2x + 10  If want to optimise the function, follow a few steps:  Step 1: Differentiate the function: f'(x) (also denoted fx or dy/dx)  Step 2: Set the differential equal to zero: f '(x) = 0.  Step 3: Solve for value of x  To establish what type of turning point(s) you have found, steps 4 & 5  Step 4: Find the second differential: f''(x) (also denoted fxx or d2y/dx2)  Step 5: Evaluate the second differential at the turning point(s).
• If f ''(x) > 0, you have a minimum point. • If f ''(x) < 0, you have a maximum point. • If f ''(x) = 0, you may have a point of inflection.

1. Multi Variables Optimisation
1.0 Differentiation
 Many relationships involve more than two variables. E.g. Demand for a good is a function of its price, advertising, price of other goods, etc  E.g. y = f(x1, x2, …, xn)  Interested in impact of change in one/all of the variables on y  Since there are more than just 1 variable, we distinguish between partial derivative and total derivative  Partial: change in y when only 1 of the x’s is changing. The partial derivative of y with respect to xi is denoted as δy/δxi or yxi or δf/δxi or fxi and is found by differentiating y with respect to xi holding all the other variables constant.  Total: change in y when all x’s are changing at the same time. It is denoted as dz/dxi.  Note difference between δ (partial derivative) & d (total derivative)

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1. Multi Variables Optimisation
1.1 Partial Differentiation
 Examples of partial differentiation Means g is a function of x1,x2,x3. Could • g(x1, x2, x3) = x13 + x22 - 3x3 δg/δx1 = 3x12 δg/δx2 = 2x2 • z = f(x,y) = 5x + δz/δx = 5 + y2 xy2 - 10 δz/δy = 2xy also have written g=f(x1, x2,x3)=…

δg/δx3 = -3

 Application: Production Functions Q = f(K,L) = 60 KL - 3K3 - 2L2 What are the marginal products of labour (MPL) and capital (MPK)? MPK = δQ/δK = 60L - 9K2 MPL = δQ/δL = 60K -4L

1. Multi Variables Optimisation
1.2 Total Differentiation
 The total differential of a function of more than one variable measures the change in the dependent variable (y) brought about by a small change in each of the independent variables (x).  If z = f (x1, x2,……xn) dz = zx1dx1 + zx2dx2 +...+ zxndxn where dx1……dxn are small changes in the independent variables.  Example: Find the total differential of z = x4 + 6xyw + 3y3 - 2w There are three variables here: x, y, w and the total differential will involve partially differentiating with respect to each of these variables: dz = (4x3 + 6yw) dx + (6xw + 9y2) dy + (6xy - 2) dw differential w.r.t.x differential w.r.t.y differential w.r.t.w

1. Multi Variables Optimisation
1.3 Second Order Partial Derivatives
 A second order partial derivative means that a function has been differentiated twice with respect to one of the independent variable while the other variables are held constant.  Consider z = f(x, y) • First order partial derivatives: fx = δz/δx fy = δz/δy • Second order: Function already been differentiated once w.r.t. x fxx = (fx)x =

  z    x  x 
And differentiated a second time w.r.t. x again Function already been differentiated once w.r.t. y

fyy = (fy)y =

  z    y  y   
And differentiated a second time w.r.t. y again

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1. Multi Variables Optimisation
1.4 Second order Cross Partial Derivatives
 There are also cross partial second order derivatives, i.e., the function is differentiated a second time, but with respect to another independent variable now, while the other variables are held constant. Function already been differentiated once w.r.t. x  fxy = (fx)y =

  z    y  x 
And differentiated a second time w.r.t. y Function already been differentiated once w.r.t. y

 fyx = (fy)x =

  z    x  y   
And differentiated a second time w.r.t. x

1. Multi Variables Optimisation
1.5 Second Order Partial and Cross Derivatives: Examples
 Consider Z = 3x2y3 + 4x3 - 5y2 zx = 6xy3 + 12x2 zy = 9x2y2 – 10y zxx = 6y3 + 24x zyy = 18x2y – 10 zxy = 18xy2 zyx = 18xy2
Function differentiated once w.r.t. x Function differentiated once w.r.t. y Function differentiated a second time w.r.t. x Function differentiated a second time w.r.t. y

Function differentiated a second time w.r.t. y Function differentiated a second time w.r.t. x

1. Multi Variables Optimisation
1.6 Optimisation of More than One Variable
 Optimisation of function with more than one variable is similar to that for a single variable.  The First order conditions determine the location of the optimal point(s). They are called the critical or turning points.  All the first order differentials must equal zero simultaneously, that is, fx1, fx2, ..., fxn = 0  The second order conditions determine the nature of that point.  An optimal point may be a maximum, a minimum, a point of inflection or a saddle point.  At the critical points, second order cross partial derivatives are equal, i.e. fxy = fyx (Young’s Theorem)

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1. Multi Variables Optimisation
1.6 Optimisation of More than One Variable..
 Consider z = f(x, y).  Following rules determine the exact nature of a turning point Maximum fx = 0 fy = 0 fxx.fyy > (fxy)2 fxx, fyy < 0 Minimum fx = 0 fy = 0 fxx.fyy > (fxy)2 fxx, fyy > 0 Saddle fx = 0 fy = 0 fxx.fyy < (fxy)2 fxx & fyy have same sign Inflection fx = 0 fy = 0 fxx.fyy < (fxy)2 fxx & fyy have different sign

 Note: if fxx.fyy = (fxy)2 test is inconclusive

1. Multi Variables Optimisation
1.6 Optimisation of More than One Variable
 How these turning points look like in a three dimensional space:
Maximum point Minimum point

Inflection point Saddle point

1. Multi Variables Optimisation
1.6 Optimisation of More than One Variable
 Example f(x, y) = x3 - 3x + xy2  First Order Differentials fx = 3x2 - 3 + y2 = 0 …………… (1) fy = 2xy = 0 …………… (2) •From (2), when x = 0, y = 0 •Replace these values in (1): when x = 0  y2 = 3 and y = +3 or -3 when y = 0  x2 = 1 and x = +1 or -1  critical/turning points are: (0, 3); (0, -3); (1, 0); (-1, 0)

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1. Multi Variables Optimisation
1.6 Optimisation of More than One Variable..
 Second order differentials: fxx = 6x fyy = 2x fxy = 2y fyx = 2y  We need to evaluate fxx, fyy, fxy and fyx at each turning point.  At (0, 3) Replace x = 0 and y =3 fxx = 0 fxx.fyy < (fxy)2 0 < 12 fyy = 0 fxx, fyy (same sign) fxy = 23  Saddle Point fyx = 23

1. Multi Variables Optimisation
1.6 Optimisation of More than One Variable
 At (0, -3) fxx = 0 fyy = 0 fxy = fyx = -23  At (1, 0) fxx = 6 fyy = 2 fxy = fyx = 0  At (-1, 0) fxx = -6 fyy = -2 fxy = fyx = 0 fxx.fyy < (fxy)2 0 < 12 fxx, fyy (same sign)  Saddle Point fxx.fyy > (fxy)2 12 > 0 fxx, fyy > 0 (6, 2 > 0)  Minimum Point

fxx.fyy > (fxy)2 12 > 0 fxx, fyy < 0 (-6, -2 < 0)  Maximum Point

1. Multi Variables Optimisation
1.7 Optimisation of More than One Variable - Application
 A firm in a perfectly competitive market sells two goods, QA and QB at a price of £10 and £8 respectively. If total costs are TC = 2QA2 + 2QAQB + QB2, what is the maximum level of profits for the firm? Use the second order conditions to check that it is a maximum.   (profit) = TR - TC TR = p.q = 10QA + 8QB   = 10QA + 8QB - 2QA2 - 2QAQB - QB2  First Order Condition: These are simultaneous equations which A = 10 - 4QA - 2QB = 0 we can solve to obtain values for QA & QB B = 8 - 2QA - 2QB = 0 QA = 1 and QB = 3 Revise your understanding of how to solve simultaneous equations   = 17  Are QA = 1 & QB = 3 turning points?

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1. Multi Variables Optimisation
1.7 Optimisation of More than One Variable
 We need to check whether this is a maximum point (at this point, profits are supposed to be maximised with these output levels)  Second Order Condition: AA = -4 BB = -2 AB = -2 BA = -2 fxx.fyy = AA.BB = 8 (fxy)2 = (AB)2 = (BA)2 = 4 fxx.fyy > (fxy)2 , i.e. 8>4 AA and BB are negative So we have a maximum point. That is, output levels QA = 1 & QB = 3 give a maximum profit level of £17.

2. Constrained Optimisation I
2.1. Constraints
 Maximising or minimising some variable is often subject to some constraint  Consider maximising a utility function
• U = 2X + 3Y • The amount of good X and Y that will maximise utility is close to infinity • But we cannot buy an infinite amount of X and Y (why?)

 Consider minimising a total cost of production function
• • • • TC = 4X1 + 6 X22 + 7X3 The minimum value will be zero with zero unit of output of each good But this is not feasible because we need to produce an output! So the output level becomes a constraint

 Therefore, most optimisation (minimisation or maximisation) problems are subject to some constraints (budget, output)

2. Constrained Optimisation I
2.1. Constraints
 Consider maximising the following utility function U = f(X, Y) subject to a given budget constraint. Price of X and Y is given as PX and PY and the consumer has an income of M.  Therefore the budget constraint is given as PXX + PYY = M  Graphically, we have:
Y

X

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2. Constrained Optimisation I
2.1. Constraints
 At equilibrium, the slope of the budget constraint is equal to the slope of the utility function (MU means marginal utility)  So

PX Y   PY X

U U

X Y

 

MU X MU Y

(sign?!)

 So Price ratio = MRS (Marginal Rate of Substitution) or

MU X MU Y  PX PY

 Ratio of MU to Price is the same for all goods

2. Constrained Optimisation I
2.1. Constraints
 Consider minimising the following Total Cost function PXX + PYY Subject to producing w amount of output Q = f(X, Y) = w

Y

Q=w X

2. Constrained Optimisation I
2.2. The Substitution Method
 How do we optimise subject to a constraint?  Use the substitution method  We substitute the constraint in the objective function and then do the maximisation procedure.  Example: Maximise Z=2x2-3xy+2y+10 subject to y –x = 0  Step 1: Rearrange constraint as y in terms of x y=x  Step 2: Substitute y = x in Z  Z = 2x2-3xx+2x+10  Z = 2x2-3x2+2x+10  Z = -x2+2x+10 This is what we seek to optimise now

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2. Constrained Optimisation I
2.2. The Substitution Method
 Step 3: Optimise Z in the conventional way Z = -x2+2x+10

dz  2 x  2  0 dx
Thus, x = 1

 When x = 1, y = 1 

d 2z  2 so turning point (1,1) is a maximum point dx 2

2. Constrained Optimisation I
2.3. The Substitution Method: Application 1
A firm faces a production function Q= 4LK + L2 and buys the inputs K and L at prices per unit of $1 and $2, respectively. If it has a budget of $105, what combination of K and L should it use in order to produce the maximum possible output? Also, verify that the ratio of Marginal product to price is the same for both factors.  Step 1: rewrite the constraint in a form for substitution 1K + 2L = 105  K = 105-2L Step 2: Substitute the constraint into the production function Q= 4L(105-2L) + L2 Q= 420L – 7L2



2. Constrained Optimisation I
2.3. The Substitution Method: Application 1
• Step 3: Optimisation of Q= 420L – 7L2

dQ  42014L  0 dL
• Therefore L = 30, and replacing L = 30 in the constraint, we can solve for K K = 105-2(30) = 45 • (30, 45) is a turning point • We check for second order condition

d 2Q  14  0 dL2
• Thus point (30, 45) is a maximum point • Replace L = 30 and K = 45 in objective function to get maximum output. • Maximum Output is hence Q= 4(30)(45) + (30)2 = 6300

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2. Constrained Optimisation I
2.3. The Substitution Method: Application 1
 MPL = δQ/δL = 4K + 2L • MPK = δQ/δK = 4L • At equilibrium:
MPL MPK  PL PK MPL 4 K  2 L 4(45)  2(30)    120 PL 2 2 MPK 4 L 4(30)    120 PK 1 1

• Thus, the ratio of Marginal Product to price is the same for both factors

2. Constrained Optimisation I
2.4. The Substitution Method: Application 2
A firm faces a Cobb Douglas production function Q= 2K1/2L1/2 and can buy the inputs K and L at prices per unit of $4 and $3, respectively. What is the cheapest way of producing 160 units of output? • Here the aim is to minimise total cost subject to producing an output of 160. The objective function is therefore the total cost function and the constraint is the production function. So we have: Minimise TC = 4K + 3L Subject to 2K1/2L1/2 =160 • Step 1: Rewrite the constraint 2K1/2L1/2 =160  L = 6400/K • Step 2: Substitution TC = 4K + 3(6400/K)  TC = 4K + 19200/K

2. Constrained Optimisation I
2.4. The Substitution Method: Application 2
• Step 3: Optimisation of TC = 4K + 19200/K

dTC 19200  4 0 dK K2
• Solving for K and L, we get K = 69.28 and L = 92.38 • Therefore (69.28, 92.38) is a turning point • Check for second order conditions:

d 2TC 38400  0 dK 2 K3

Note: we replace these values in the original cost function, i.e. TC=4K+3L

• So it is a minimum point. • Replacing these values in the TC function, the minimum cost of producing 160 units of output is given as $554.26

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3. Constrained Optimisation II
3.1. The Lagrangian Multiplier (LM) Approach
 Substitution Method has certain disadvantages
• Becomes more complicated when deal with more than 2 variables • Does not provide additional information from maximisation/minimisation process

 Lagrange Multiplier Approach overcomes these problems, and hence:
• Makes it easier to deal with more than 2 variables • Enables us to obtain additional information from the maximisation/minimisation process, namely the Langrage multiplier ()

3. Constrained Optimisation II
3.1. The Lagrangian Multiplier (LM) Approach
 If we want to optimise an objective function, f(x1...xn) subject to a constraint, g(x1...xn) = M, the LM approach proceeds as follows:  Step 1: Define the Lagrangian function • L(x1..xn, ) = f(x1....xn) + [M - g(x1...xn)] where  is known as the Lagrange multiplier.  Step 2: Find the partial derivatives, Lx1,...Lxn, L • Set partial derivatives equal to zero and solve for x1...xn and .  Step 3: Check for second order conditions  The Lagrange Multiplier () can be thought of as the effect on the objective function of a unit change in the constraint (marginal effect)

3. Constrained Optimisation II
3.2. The Lagrange Multiplier Approach: Example
 Optimise f(x, y) = x2 - 3xy + 12x subject to 2x + 3y = 6  Step 1: Set up the Lagrangian function L = x2 - 3xy + 12x + (6 - 2x - 3y)  Step 2: Find the partial derivatives Lx = 2x - 3y + 12 - 2= 0 ……. (1) Ly = -3x -3 = 0 ……. (2) L = 6 - 2x -3y = 0 ……. (3)  From (2):  = -x  Substituting (4) into (1): 2x -3y + 12 + 2x = 0 3y = 4x + 12 ……. (4)

……. (5)

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3. Constrained Optimisation II
3.2. The Lagrange Multiplier Approach: Example
 Substituting (5) into (3): 6 - 2x - 4x -12 = 0 -6x = 6 x = -1 ……. (6)  Substituting (6) into (5): y = 8/3  Substituting (6) into (4): =1   = 1 implies that a unit change in the constraint will increase the objective function by 1.

3. Constrained Optimisation II
3.3. The Lagrange Multiplier Approach: Application
 Suppose a utility function is given by: U = 40x0.25y0.5 with prices of x and y as px = 4, py = 10 and income M = 60 What level of x and y will maximise utility? What is the meaning of the Lagrange multiplier in this case?  Budget contraint: 4x + 10Y = 60  Lagrange Function: L = 40x0.25y0.5 + (60 - 4x - 10y)  First Order Conditions Lx = 10x-0.75 y0.5 - 4 = 0  2.5x-0.75 y0.5 =  Ly = 20x0.25y-0.5 - 10 = 0  2x0.25y-0.5 =  L = 60 - 4x - 10y = 0 ….

… (1) … (2) (3)

3. Constrained Optimisation II
3.3. The Lagrangian Multiplier Approach: Application
 Equating (1) and (2) 2.5x-0.75 y0.5 = 2x0.25y-0.5 4x = 5y …… (4)  Substituting (4) into (3): 60 - 5y - 10y = 0 60 = 15y y=4  From (4) x=5  From (1) or (2)  = 1.495  1.5

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3. Constrained Optimisation II
3.3. The Lagrangian Multiplier Approach: Application
What is the meaning of ? If we relax the budget constraint by a small amount, utility will increase by  times that amount. So if we give the consumer an extra £1 income, utility will go up by about 1.5 units. It is also called the marginal utility of income.  Note that we could have set up the Lagrangean function as follows: L = 40x0.25y0.5 + (4x + 10y - 60)  We would still get same answer for x and y, but  = -1.5  The interpretation of  is not appropriate (the consumer’s total satisfaction should increase, and not decrease if you give him more income!)  Therefore, should be careful when defining the Lagrangian function!

4. Exercises
 Exercise 1
The total cost (TC) & total revenue (TR) functions for a company are as follows: TC = 2q2 +2a.q + 2a2 TR = 18q + 12a + a.q where q is the quantity of output and a is the expenditure on advertising. (i) Write down the profit function for this firm. (ii) Find the profit maximising levels of output and advertising. (iii) Find the maximum level of profits

4. Exercises
 Exercise 2
The output function of a firm is given by Q = 120L + 200K – L2 – 2K2 where L = quantity of labour and K = quantity of capital. Unit labour cost is £5 and unit capital cost is £8. (i) Use the Lagrangian method to find the maximum level of output the firm can produce with a budget of £130. (ii) Explain the meaning of the Lagrangian multiplier in this example.

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4. Exercises
 Exercise 3
A consumer has the following utility function over two goods, X and Y: U = 20 X0.5Y0.5 (i) If the price of X is 20, the price of Y is 5 and the consumer’s income is 300, find the maximum value of utility that the consumer can achieve. (ii) Calculate the value of the Lagrangian multiplier and explain its meaning.

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