Chapter 4_6.pptx

Chapters 4.6

Optimization

Optimization

is using calculus to either

•

minimize

an undesirable quantity, such as

cost, or

•

maximize

a desirable quantity, such as

profit, revenue or demand.

In this section we pursue optimization of

narrative problems.

First, however, we would like to review some

derivative rules and examples.

Derivatives of functions requiring arguments:

Example:

Example:

Generalized power rule:

If

then

Product Rule:

Quotient Rule:

In Chapter 4.4 we indicated that a

point of

diminishing returns

occurs at the

inflection

point

of an increasing function which changes

concavity from upwards to downwards, as shown

in the graph below.

At the inflection

is

increasing most

rapidly

and the rate of

return for the investment

is

optimized

.

S(a)

Problem 1:

Telemarketing is the major means of advertising

for a small business. The following table provides

the monthly profit of the business, in thousands of

dollars, as a function of monthly labor hours

devoted to telemarketing.

telemarketing labor hours

profit ($Th) telemarketing

labor hours

profit ($Th)

0 2 50 43.0

10 3.5 60 48.5

20 8.5 70 55.5

A. Scatterplot the telemarketing data. You should

find that a cubic model would be appropriate.

Determine the cubic model, reporting the model

parameters and R

to 5 decimal digits.

B. Find the 1

and 2

derivative of the cubic

model.

C. For what number of telemarketing labor hours is

profit increasing most rapidly?

D. What is the profit associated with your result

from Part C?

Problem 1 Solved:

A.

Problem 1 Solved (cont):

Problem 1 Solved (cont):

an inflection point occurs at x = 39.3

Since the model is an increasing function, the inflection point indicates the input value at which the output is

increasing most rapidly. That is, when 39.3 labor hours are invested in telemarketing, profit is increasing most rapidly.

This is the point of diminishing returns. Investing more hours into telemarketing will generate more profit, but at a diminishing rate.

Problem 1 Solved (cont):

D. an inflection point occurs at x = 39.3

Monthly profit is approximately $29,290 when 39.3 hours are invested in telemarketing monthly.

Problem 1 Solved (cont):

E. an inflection point occurs at x = 39.3

Profit is increasing at approximately $1,060 per hour when 39.3 hours are invested in telemarketing monthly.

Consider:

Market share

is the percentage of a market (defined in

terms of either units or revenue) accounted for by a

specific company or organization. Market share is a

key indicator of market competitiveness—that is, how

well a firm is doing against its competitors.

The market share for a company’s product, , months

since its introduction into the marketplace, can be

modeled with the following function during the first year

after its introduction:

Problem 2:

A.

Find the first derivative of the market share

model.

B.

Find the second derivative of the market

share model.

A.

Recall the quotient rule:

B. Starting with the first derivative and using the quotient rule:

Problem 2 Solved:

(

� ₂

)

′

(�)= �1

′

(�) � ₂(�)− �₂′(�) �₁(�)

Problem 2:

C.

Does the model reach a relative extreme

over the interval 0



t



12, that is, during

the first year.

D.

When did the product reach its maximum

market share during the first year and

C.

  relative extreme occurs at months

D. candidate input values at which a maximum market share occurs on the interval :

t = 3, 0, 12

maximum market share is 22%, and occurs 3 months after product introduced

Problem 2 Solved (cont):

Problem 2:

E.

Graph the market share function over the

first year.

F.

Your graph should indicated the

rate of

change

of the market share reaches a

maximum negative value around 4 to 6

months after introduction. Thereafter, the

rate of change begins to slow. Determine

the time when the rate of change of

E. See graph below F. 

using the solver, , 5.2, -5.2

Decline in market share begins to slow 5.2 months after introduction of the product.

Problem 2 Solved (cont):

Consider:

A Greyhound bus route currently transports, on

average, 8000 passengers per month at a fare of $50.

Market research indicates the company will lose 100

monthly passengers for each $1 increase in fare.

We wish to determine the fare which maximizes

revenue.

The number of passengers per month, N, and the

revenue, R, will be dependent upon the fare. Therefore

identify:

The statement “the company will lose 100 monthly

passengers for each $1 increase in fare” indicates the

rate of change

in number of passengers, with respect

to the fare, will be .

This is a

constant rate of change

and indicates the

relationship between number of passengers and the

fare is

linear

. That is, a straight line can express the

relationship between fare and number of passengers:

Recall:

x = fare per passenger ($/passenger)

N(x) = number of passengers per month

So far we know

The statement “8000 passengers per month at a fare of

$50” gives an ordered pair:

Solving for b:

The revenue, , is the fare/passenger, , times the

number of passengers, .

Revenue may be maximized for the fare,

, which yields

a zero derivative and negative 2

derivative:



Problem 3:

A small ski resort offers an end-of-season $200 weekend special providing it can sell at least 100 reservations. The maximum

number of reservations is the facility capacity of 160. For each reservation in excess of 100, the price will be decreased by $1 per reservation.

We wish to determine:

• the number of reservations which maximizes revenue, and • the maximum revenue.

The price per reservation, P, and the revenue, R, are dependent upon the number of reservations. Identify:

x = number of reservations

P(x) = price per reservation ($/reservation) R(x) = revenue ($)

Problem 3 (cont):

The price per reservation, P, and the revenue, R, are dependent upon the number of reservations. Identify:

x = number of reservations

P(x) = price per reservation ($/reservation) R(x) = revenue ($)

Note:

1. We are considering price, , as a function of the number of reservations, .

2. Revenue is the product of the number of reservations and the price per reservation, that is,

3. Note that in this problem we are constrained by 100 ≤ x ≤ 160.

Problem 3 (cont):

A. The statement “For each reservation in excess of 100, the price will be decreased by $1 per reservation” indicates the

rate of change in price per reservation, with respect to the

number of reservations, will be . This is a constant rate of

change and indicates the relationship between number of

reservations and the price is linear. That is, a straight line can express this relationship between the input variable, x, and the output variable, P(x). Write a generic linear model relating these two variables.

B. What is the value of the slope of this linear relationship?

C. The statement “$200 weekend special providing it can sell at least 100 reservations” gives an ordered pair: . Use this

statement to find the vertical axis intercept, b, of the linear model.

A.

B. 

C. one ordered pair: Solving for b:

D.

Problem 3 (cont):

E. Write the equation for R(x).

F. Does R(x) reach a relative extreme on the interval 100 ≤ x ≤ 160?

G. Determine the 2nd derivative of the revenue function and indicate what it tells us about any relative extreme values.

H. Evaluate all candidate input values to determine the maximum revenue.

E.

F. 

G.  any relative extreme is a max

produces a relative maximum revenue

Problem 3 Solved (cont):

H. Candidate number of reservations, x, which may maximize revenue on the interval 100 ≤ x ≤ 160:

produces a maximum revenue of $22,500

a price of $150 maximizes revenue.

Problem 3 Solved (cont):

Consider:

Frozen orange juice sells consistently thru the year. A

supermarket wishes to optimize the ordering of OJ by

submitting orders of the same size thru the year. It is

estimated annual sales will be 1200 cases.

The manager is concerned with two competing costs:

First, each delivery will cost $75, so less deliveries per

year reduces the delivery costs.

However, the storage cost per case is $8 per year

where the average number of cases stored per year is

50% of the average number of cases per order. Less

deliveries means more cases per order, increasing

Set

= # of annual orders

Since we require 1200 cases annually, the number of

cases per order is:

= # cases/order

Delivery costs, , are a function of the # of orders,

specifically, $75 per order:

Storage costs, , are a function of the # cases per

order, . Specifically, 50% of the cases per order, are

expected to be stored per year at an annual cost of $8

per case:

Problem 4:

A.Write the total cost function, .

B.Determine the number of orders, , which minimizes

the total cost function.

A. , x = # of annual orders

B.

 

which is always greater than zero

Therefore produces a relative minimum cost C.

Problem 5:

A tire manufacturer has annual orders for 600,000 for a specific tire model. The production costs, C_p, to set up the plant for each model are $15,000 per production run.

Storage cost, C_s, per tire per year are $5. Assume the average number of tires stored per year is 50% of the average number of tires per run.

Again, less production runs reduce production costs at the expense of storage costs.

Let = # of tires produced per run

= # of tires produced per run

Since we require 600,000 tires annually: = # runs

Production costs, , are a function of the # of runs, specifically, $15,000 per run:

Storage costs, , are a function of the # tires per run, .

Specifically, 50% of the tires per run, are expected to be stored per year at an annual cost of $5 per tire:

A.

 

> 0 therefore the x value generates a relative minimum cost B.

C.

Problem 6:

A museum experiments with the Sunday ticket price,

generating the following data.

ticket price ($)

average # Sunday

customers

20

1600

25

1500

30

1250

32

850

A.Scatterplot the data and find the best model

which

expresses S(x), that is, # of customers as a function

of ticket price, x.

B.Write an expression for revenue, R(x), as a

function of ticket price, x. Note, R(x) = x S(x).

C.Determine the ticket price which maximizes

revenue over the interval $20 ≤ x ≤ $32.

D.The museum estimates that the average total cost

per customer to operate the facility is $10. Write an

expression for cost, C(x), as a

function of ticket

price, x. Note, C(x) = $10 S(x).

A. B.

C.

 relative minimum revenue

 relative maximum revenue

Max revenue on interval $20 ≤ x ≤ $32:

D. E.



 relative min

 relative max

E. Max profit on interval $20 ≤ x ≤ $32:

The ticket price which maximizes profit, $27.58, is a bit different than that which maximizes revenue, $26.05.