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Chapter two

Optimal Decisions Using Marginal Analysis

Managerial Economics, 8e

William F. Samuelson ● Stephen G. Marks

RECAP

Chapter 2: Optimal Decisions Using Marginal Analysis

U = number of actual users

M = number of potential users

a = % of actual users lost in each period

b = % of potential users gained in each period

The sustainable user base is:

RECAP

Chapter 2: Optimal Decisions Using Marginal Analysis

= cost of reducing a by 1%

= cost of increasing b by 1%

The optimal allocation of the marketing budget is:

Calculate the derivatives and find the optimal ratio

RECAP

Chapter 2: Optimal Decisions Using Marginal Analysis

The derivatives are:

EXAMPLE 1: MICROCHIP MANUFACTURER

Demand:

Revenue:

Chapter 2: Optimal Decisions Using Marginal Analysis

Other factors than price that may impact demand are held fixed. They would affect the parameters.

Sales are lots (of 100 chips) per week.

Price per lot is in thousands of dollars.

Figure 2.1 The demand curve for microchips

Chapter 2: Optimal Decisions Using Marginal Analysis

Figure 2.2 revenue from microchips

Chapter 2: Optimal Decisions Using Marginal Analysis

EXAMPLE 1:

Cost:

Profit:

Chapter 2: Optimal Decisions Using Marginal Analysis

Quadratic function that increases, then decreases.

Cost is in thousands of dollars.

Fixed cost: $100K per week. Variable cost: $38K per lot.

Figure 2.3 The cost of microchips

Chapter 2: Optimal Decisions Using Marginal Analysis

Figure 2.4 Profit from microchips

Chapter 2: Optimal Decisions Using Marginal Analysis

Figure 2.5 A close-up view of profit

Chapter 2: Optimal Decisions Using Marginal Analysis

EXAMPLE 1:

Profit:

Chapter 2: Optimal Decisions Using Marginal Analysis

Derivative of Q) is the slope of the profit function at Q. Slope is zero at that maximizes profit.

Recall the profit function.

First-order condition:

→

Figure 2.6 Total profit and marginal profit

Chapter 2: Optimal Decisions Using Marginal Analysis

DIFFERENTIATION

Chapter 2: Optimal Decisions Using Marginal Analysis

Suppose we have a function of the form:

The derivative is:

Special cases:

DIFFERENTIATION

Chapter 2: Optimal Decisions Using Marginal Analysis

For instance, for the function:

Derivatives of the additive terms are:

DIFFERENTIATION

Chapter 2: Optimal Decisions Using Marginal Analysis

The derivative of a sum of terms is the sum of the derivatives of the additive terms:

So:

EXAMPLE 2:

Profit:

Chapter 2: Optimal Decisions Using Marginal Analysis

Derivative of Q) is the slope of the profit function at Q. Slope is zero at that maximizes profit.

Suppose we have a quadratic profit function.

First-order condition:

→

Figure 2A.1 The Firm’s Profit Function

Chapter 2: Optimal Decisions Using Marginal Analysis

EXAMPLE 3:

Chapter 2: Optimal Decisions Using Marginal Analysis

Consider hypothetically a cubic profit function:

It has two flat points, where its derivative is zero. One is a minimum, one is a maximum.

Profit:

Figure 2A.2 A Second Profit Function

Chapter 2: Optimal Decisions Using Marginal Analysis

EXAMPLE 3:

Chapter 2: Optimal Decisions Using Marginal Analysis

Recall the cubic profit function:

Solutions of the first-order condition are Q = 2 and Q = 10. For a maximum, we also need > 6:

First-order condition:

Second-order condition:

MAXIMA AND MINIMA

Chapter 2: Optimal Decisions Using Marginal Analysis

Near a maximum, a function rises, then falls. Its derivative changes from positive to negative and keeps declining. The derivative of its derivative is negative.

Near a minimum, a function falls, then rises. Its derivative changes from negative to positive and keeps increasing. The derivative of its derivative is positive.

EXAMPLE 4: MULTIVARIATE OPTIMIZATION

Profit:

Chapter 2: Optimal Decisions Using Marginal Analysis

Maximize separately with respect to P and A.

Suppose the firm simultaneously decides on pricing and an advertising budget.

First-order conditions:

EXAMPLE 4:

Chapter 2: Optimal Decisions Using Marginal Analysis

Solve jointly by substituting one into the other.

Recall the dual first-order conditions.

EXAMPLE 5: CONSTRAINED OPTIMIZATION

Profit:

Chapter 2: Optimal Decisions Using Marginal Analysis

The unconstrained optimum (Q = 5) does not satisfy the constraint. The derivative of the profit function is negative in the allowed range (Q ≥ 7):

Given a quadratic profit function, suppose that a minimum output of Q ≥ 7 is required.

Therefore, Q should be as small as possible: = 7.

EXAMPLE 6: CONSTRAINED MULTIVARIATE OPTIMIZATION

Profit:

( )

Chapter 2: Optimal Decisions Using Marginal Analysis

Assume a quadratic profit function in two outputs, subject to the production constraint + ≤ 25.

The unconstrained optimum ( = 20, = 20) violates the constraint. It is optimal to equate the derivatives of the profit function with respect to a, and to make the constraint bind:

EXAMPLE 6:

Chapter 2: Optimal Decisions Using Marginal Analysis

Recall the optimality conditions.

Solve jointly through substitution:

EXAMPLE 6:

Chapter 2: Optimal Decisions Using Marginal Analysis

At the constrained optimal solution ( = 10, = 15), the incremental profit from increasing each output is the same.

If it were not so, the firm could raise profit by increasing the output with the higher return while decreasing the other to satisfy the constraint.

PARTIAL AND TOTAL DERIVATIVES

Chapter 2: Optimal Decisions Using Marginal Analysis

The partial derivative ∂𝜋()/ ∂ only takes the direct effect of on 𝜋() into account, while keeping fixed.

The total derivative d𝜋()/ d accounts for the full impact of on 𝜋(), including indirect effects from changes in .

They are the same if d / d = 0. They are distinct when and are linked by a constraint (so that must be reduced when is increased).

THE CHAIN RULE

Chapter 2: Optimal Decisions Using Marginal Analysis

When differentiating a composite function f(g(x)) with respect to x, the derivative is sequentially “unpacked,” and the resulting terms multiplied, to capture the impact of x on f(g(x)), first through the direct effect on g(x), then through the effect of g(x) on f(g(x)).

For instance, the constraint + = 25 implies = 25 – = g(), and 𝜋() = f(, g()). We can use the chain rule to determine how an increase in , while reducing as necessary, affects profit.

THE CHAIN RULE

Chapter 2: Optimal Decisions Using Marginal Analysis

Specifically, recall:

The total derivative with respect to is:

( )

Set to zero, this gives the right solution for .

THE PrINCIPLE OF OPTIMALITY

Chapter 2: Optimal Decisions Using Marginal Analysis

Without constraints on the variables, profit is maximized by setting the derivative of the profit function to zero.

With constraints on the variables, profit is maximized by setting derivatives with respect to each variable equal.

The ratio d is imposed by the constraint. (In Example 6, it was –1, because increasing by one unit required reducing by one unit.)

THE PrINCIPLE Of OPTIMALITY

Chapter 2: Optimal Decisions Using Marginal Analysis

We can obtain the constrained principle of optimality from the unconstrained principle if we apply the first-order condition to the total derivative:

THE PrINCIPLE OF OPTIMALITY

Chapter 2: Optimal Decisions Using Marginal Analysis

Since

the unconstrained principle implies:

I.e., profit is maximized when marginal revenue equals marginal cost.

THE PrINCIPLE OF OPTIMALITY

Chapter 2: Optimal Decisions Using Marginal Analysis

The constrained principle is a special case of the marginal-benefit-equals-marginal-cost rule. Return to is the opportunity cost of increasing when there is a constraint on total output.

Marginal benefit of raising is profit gained from higher :

Marginal cost of raising is profit lost from lower :

EXAMPLE 1:

Chapter 2: Optimal Decisions Using Marginal Analysis

Revenue is therefore:

Recall that, for the microchip manufacturer, demand was:

Marginal revenue is the derivative with respect to Q:

This is the increase in revenue from selling another unit.

EXAMPLE 1:

Chapter 2: Optimal Decisions Using Marginal Analysis

The microchip manufacturer’s cost function was:

Marginal cost is the derivative with respect to Q:

This is the increase in cost from selling another unit.

EXAMPLE 1:

Chapter 2: Optimal Decisions Using Marginal Analysis

Profit is maximized when MR = MC:

This solution is the same as that obtained from the first-order condition d𝜋(Q)/ dQ = 0.

Profit-maximizing price and maximal profit follow from .

THE FIRM’S OPTIMAL OUTPUT DECISION

The Firm determines Output

where MR = MC.

0 2 4 6 8

R, C



R = 170Q - Q2

300

200

100

-100

C = 100 + 38Q



3.3

M = 0

Chapter 2: Optimal Decisions Using Marginal Analysis

Figure 2.7 Marginal revenue and marginal cost

Chapter 2: Optimal Decisions Using Marginal Analysis

LINEAR DEMAND

Chapter 2: Optimal Decisions Using Marginal Analysis

Suppose (inverse) demand has the linear form:

Revenue is:

Then marginal revenue is:

Marginal revenue associated with linear demand has the same intercept and double the slope as the inverse demand function.

Maximum

Contribution

MAXIMIZING PROFIT USING MARGINAL GRAPHS

Set MR = MC.

170

Demand

Chapter 2: Optimal Decisions Using Marginal Analysis

SPREADSHEET ANALYSIS

Chapter 2: Optimal Decisions Using Marginal Analysis

To download the Solver Add-In in Excel:

1. File → Options → Add-Ins

2. Manage: Excel Add-Ins → Go

3. Select Solver Add-In → OK

You’ll find it on the Data tab, in the Analysis group of tools.

Table 2A.1 Optimizing a Spreadsheet

Chapter 2: Optimal Decisions Using Marginal Analysis

Table 2A.2 Optimizing a spreadsheet

Chapter 2: Optimal Decisions Using Marginal Analysis

Figure 2.8 Shifts in mr & Mc

Chapter 2: Optimal Decisions Using Marginal Analysis

SENSITIVITY ANALYSIS

Considers changes in: Fixed Costs, Marginal costs,

or Demand Conditions

170

Demand

A change in fixed cost has no

effect on Q* or P* (because

MR and MC are not affected).

Chapter 2: Optimal Decisions Using Marginal Analysis

SENSITIVITY ANALYSIS

Considers changes in: Marginal costs

170

Demand

Q’

MC’

An increase in MC

implies a fall in Q

and an increase in P.

Chapter 2: Optimal Decisions Using Marginal Analysis

SENSITIVITY ANALYSIS

Finally, consider a change

in Demand Conditions.

170

Shift in

Demand

Q

P

The favorable demand shift

calls for an increase in Q

and an increase in P.

Chapter 2: Optimal Decisions Using Marginal Analysis

APPLICATION 1: PRICING AMAZON KINDLE

Chapter 2: Optimal Decisions Using Marginal Analysis

Data Points:

At $259, 1 million Kindles were sold per year

Sales tripled when price was cut to $189

Each Kindle sells about 25 e-books that, on average, generate $4 profit (i.e. $100 profit per Kindle)

Estimated cost of producing a Kindle is $126

Marginal cost:

Modeling demand (in million Kindles):

Reflects that P = 259 at Q = 1 and P = 189 at Q = 3.

APPLICATION 1:

Chapter 2: Optimal Decisions Using Marginal Analysis

If Amazon treated Kindle sales independently from e-book sales, it would set marginal cost of $126 equal to marginal revenue:

Profit maximization would seem to require:

APPLICATION 1:

Chapter 2: Optimal Decisions Using Marginal Analysis

This suggests that the Kindle should be priced even lower.

If Amazon recognizes the additional $100 revenue from e-book sales per Kindle, the marginal cost of a Kindle is effectively only $26. Then profit is maximized if:

APPLICATION 2: FRANCHISE CONTRACTS

Chapter 2: Optimal Decisions Using Marginal Analysis

Facts:

Franchisors earn a percentage of revenue

They focus on marketing and quality control to build and protect the brand

Franchisees optimize operations to sell at the lowest possible cost

Conflicts of interest:

Franchisors aim to maximize store revenue (MR = 0)

Franchisees aim to maximize store profit (MR = MC > 0)

Franchisees often oppose discounts they have to give and investments in service they have to pay for