optimal decisions using margial analysis

8/6/2019 Optimal Decisions Using Margial Analysis

1/43

Chapter 2

Optimal Decisions UsingMarginal Analysis


2/43

What is economic

optimization? If there is only one solution to a

problem, then, there is no decisionproblem.

If there are alternative solutions, thenyou have to decide about the bestsolution.

The best (optimal) solution is one thatproduces the result most consistentwith the managerial objectives.


3/43

Economic Optimization Economic optimization is the

process of arriving at the best

managerial decision. Managerial economics provides

tools for analyzing and evaluating

decision alternatives.


4/43

Primary Objective of

Management In managerial economics, the primary

objective of management is assumed to

bemaximizing the value of the firm

Maximize the following equation:

n

n

2

2

1

1

k)(1

CF...

k)(1

CF

k)(1

CFV

+

++

+

+

+

=


5/43

Total, Average, and

Marginal Relations In any mathematical relationship, there

is a DEPENDENT VARIABLE and one or

more INDEPENDENT VARIABLE(S). A marginal relation is the change in the

dependent variable caused by a one-unit change in an independent variable.


6/43

Total, Average, and

Marginal Relations

Units of Ou

Q


7/43

0

50

100

150

200

250

0 2 4 6 8 10

Output (un

Profits($)

Total Prof its 0 Marg ina l Prof its 0 Av erage Prof


8/43

Total, Marginal, and

Average Relations When the marginal is positive, the

total is increasing; when the

marginal is negative, the total isdecreasing.

Maximization of the total profit

function occurs at the point wherethe marginal switches frompositive to negative.


9/43


Average Relations When the marginal is greater than the

average, the average must beincreasing.

E.g. You are operating 5 stores withaverage annual sales ofTL 100 billionper store. If you open a 6th store that

generates sales ofTL 120 billion (themarginal sales), average sales perstore will increase.


10/43


Average RelationsTotal Output is denoted by Q andTotal Profit is denoted by .

Marginal profit is expressed as where denotes difference orchange.

= Q - Q-1

Average profit equals total profitdivided by total output ( / Q).


11/43


Average Relations Marginal profit is the slope of the

total profit curve; it is maximized at

the inflection point.Total profit is maximized where

marginal profit is equal to zero.

Average profit rises (falls) whenmarginal profit is greater (less) thanaverage profit.


12/43

Marginals as Derivatives of

Functions In function Y = f(X), the marginal

value is the change in Y associated

with a one-unit change in X.Then, the general specification is

Marginal Y = Y / X.

A derivative is the precisespecification of the marginalrelation.


13/43


Functions A derivative is the value ofY/ X

for extremely small changes in X.

The derivative is also equivalent tothe slope of a curve at a given point.

The derivative is denoted as dY/dX.

Since the derivative is the slope of acurve, it will change as the slope ofthe curve changes.


14/43


Functions E.g. When the dependent variable

is Total Profit (Y) and the

independent variable is TotalOutput (X), the derivative will showthe exact mathematical relation

between and Q at a specificlevel of Q.


15/43

Marginal AnalysisTo find the maximum or minimum

value of a function, we need to find the

point where the marginal value is equalto zero.

When the derivative is used to measurethe marginal, the function will bemaximized or minimized when thederivative is equal to zero.


16/43

Marginal Analysis

E.g. = -10,000 + 400 Q - 2 Q2

Marginal Profit (M ) = d /dQ

d /dQ = 400 - 4Q Set the derivative equal to zero:

400 - 4Q = 0 Q = 100 units

Total profit is maximized at an outputlevel of 100 units.


17/43

Maximum and MinimumValues

The derivative is zero for both amaximum and a minimum point.

The second derivative is used todistinguish maximums from minimums.

The second derivative gives the slope ofthe slope (shows the change in the slopeof the original function).


18/43

The Second Derivative

If the second derivative isnegative, this indicates a

maximum point on the originalfunction.

The negative second derivative

shows that the marginal function ischanging from a positive slope to anegative slope.


19/43

The Second Derivative

If the second derivative is positive,this indicates a minimum point on

the original function.The positive second derivative

shows that the marginal function is

changing from a negative slope toa positive slope.


20/43

Marginal Profit

Total Profit

A

B

A B

Inflection Point

Profitpertime

period

Profi

outp

Units of output

Units of output


21/43

Multivariate Optimization

It is the process of optimization forequations with three or more

variables. E.g. Demand is a function of (1)

products own price, (2) price of

other goods, (3) advertising, and(4) income.


22/43

Partial Derivatives

Optimization requires an analysis ofhow a change in each independent

variable affects the dependent variable,holding constant the effect of allother independent variables.

Partial derivatives are used for finding

this type of an isolated effect.


23/43

Partial Derivatives

E.g. Q = 5000 - 10P + 40A + PA - 0.8A2 -0.5P2

In order to find the effect of eachindependent variable, we will assume thatall variables except the one under analysisremain unchanged (constant). Othervariables will be treated as constants whentaking the partial derivatives.


24/43

Partial Derivatives

Q = 5000 - 10P + 40A + PA - 0.8A2 -0.5P2

Q/ P = 0 - 10 + 0 + A - 0 - P = -10 +A - P

Q/ A = 0 - 0 + 40 + P - 1.6A - 0

= 40 + P - 1.6A


25/43

Maximizing MultivariateFunctions

All first-order partial derivatives areset equal to zero to find the

maximum of a multivariate function. Q/ P = -10 + A - P = 0

Q/ A = 40 + P - 1.6A = 0

P = 40, A = 50, Q* = 5,800


26/43

E.g. TC = 200,000 + 30Q + 0.002Q2

TR = 250Q - 0.02Q2

Q = 12,500 - 50P

Calculate the revenue-maximizingprice/output combination and the profitat this level.

dTR/dQ = 250 - 0.04Q Q = 6,250

P = 125

= TR - TC = -200,000 + 220Q -

0.022Q2

= 315,625


27/43

Calculate the profit-maximizingprice/output combination and the profitat this level.

= -200,000 + 220Q - 0.022Q2

d /dQ = 220Q - 0.044Q Q = 5,000

P = 150

= 350,000


28/43

Calculate the average cost-minimizingoutput level and the profit at this level.

TC = 200,000 + 30Q + 0.002Q2

AC = TC/Q = 200,000Q-1 + 30 + 0.002Q

dAC/dQ = -200,000Q-2 + 0.002

Q = 10,000

P = 50

= -200,000 (loss)


29/43

Revenue-maximizing and profit-maximizingresults show that revenue-maximizingprices are lower (demand curve slopes

downward) and output levels are higherthan the case of profit-maximization.

Profit-maximizing and average cost-minimizing results show that the output

level where average costs are minimized isnot necessarily the point where profits aremaximized.


30/43

Constrained Maximization In many decisions, there are constraints

imposed that limit the options availableto the decision maker.

E.g. Minimizing cost while producing agiven level of output.

E.g. Maximizing output while utilizing a

given level of inputs. E.g. Maximizing sales with a fixed

advertising budget.


31/43


32/43

Constrained Optimization

If the constraint can be expressed as anequation, then it can be included in the

optimization process in the followingway: Solve the constraint for one of the

independent variables

Substitute that variable in the objectivefunction that the firm wants to maximize orminimize.


33/43

Constrained Maximization

E.g. minimize TC = 3X2 + 6Y2 - XY

subject to X + Y = 20.

From the constraint X = 20 - Y. Substituting:

TC = 3(20 - Y)2 + 6Y2 - (20-Y)Y

TC = 1200 - 140Y + 10Y2


34/43

Minimize this objective function by

taking derivatives: TC = 1200 - 140Y + 10Y2

dTC/dY = -140 + 20Y = 0 Y = 7

Check the second derivative to see thatthis is a minimum:

d2TC/dY2 = 20 (it is positive)

Substitute for X: X = 20 - Y X = 13.

TC = 3(13)2 + 6(7)2 - 3(7) = 710.

The total cost of producing thiscombination is $710.


35/43

Constrained MaximizationLagrangian Multipliers

When the constraints are too complex ornumerous, the substitution method is notfeasible.

The Lagrangian method creates a new objectivefunction by incorporating the original objectivefunction and all the constraint conditions.

When the new Lagrangian function is maximized

or minimized, the original objective function ismaximized or minimized and all the constraintsare satisfied.


36/43

Lagrangian Multipliers

E.g. minimize TC = 3X2 + 6Y2 - XY

subject to X + Y = 20.

Step 1: Rearrange the constraint:0 = 20 - X - Y

Step 2: Multiply the constraint by theunknown factor and add the result tothe original objective function to createthe Lagrangian function:


37/43

LTC = 3X2 + 6Y2 - XY + (20 - X - Y)

Step 3: Treat the Lagrangian function as anunconstrained optimization problem since it

already incorporates the constraint. Step 4: Solve the optimization problem by the

use of partial derivatives:

LTC / X = 6X - Y -

LTC / Y = 12Y - X -

LTC / = 20 - X - Y


38/43

Step 5: Set the partial derivatives equal tozero and solve the system of equations:

6X - Y - = 0

12Y - X - = 0

20 - X - Y = 0

(three equations and three unknowns)

X = 13, Y = 7, TC = 710.

Th L i M lti li


39/43

The Lagrangian Multiplier( )

The Lagrangian multiplier has aneconomic interpretation:

From the example, = 71. Since theoriginal objective function is a total costfunction, is interpreted as themarginal cost of producing 20 units of

output. So, if the firm could produce 19 units, TC

would decrease by approximately 71. Or, ifthe firm could produce 21 units, TC would

increase by approximately 71.


40/43

E.g. maximize = -10,000 + 400Q - 2Q2

subject to 4Q = 300 Rearrange and write out the Lagrangian

function:L

= -10,000 + 400Q - 2Q2 + (300 - 4Q)

Take the partial derivatives and set them equalto zero:

L

/ Q = 400 - 4Q - 4 = 0

L

/ = 300 - 4Q = 0

Q = 75, = 25, = 8,750. If the right-hand-side of the constraint can be

increased by one unit, the profits will increaseby 25.


41/43

E.g. TR = 20T + 5N + 20TN - T2

where

T = units of television advertising

(cost = $10 per unit)

N = units of newspaper advertising

(cost = $5 per unit)

There is a $100 advertising budgetconstraint.

Calculate the revenue-maximizingcombination of television and newspaper

advertising.


42/43

Maximize TR = 20T + 5N + 20TN - T2

subject to 10T + 5N = 100

Rearrange the constraint and write out the Lagrangian function:

LTR = 20T + 5N + 20TN - T2 + (100 - 10T - 5N)

Take the partial derivatives and set them equal to zero:

dLTR /dT = 20 + 20N - 2T - 10 = 0

dLTR /dN = 5 + 20T - 5 = 0

dLTR /d = 100 - 10T - 5N = 0


43/43

Solve the system of equations:

T = 5 units, N = 10 units, TR = $1,125

Interpret the value of the Lagrangian multiplier:

= $21

Lambda measures the marginal revenue to begained by increasing the advertising budget by$1. It is not the revenue gain associated with aunit increase in either television or newspaperadvertising, but rather the revenue increasefollowing a $1 increase in expenditures on anappropriate combination of television and

optimal decisions using margial analysis

Documents