Section 2.2
For Problems 8-9, formulate mathematical models, and then solve the problems.
8. Premium loam is 60% soil, 40% domestic manure, and costs $5 per 50 lb. Generic
loam is 20% soil, 10% domestic manure (an 70% sand, stone, etc.), and costs $1
per 50lb. We need loam for our backyard that is at least 36% soil and at least 20 %
domestic manure. What combination of the two loams should we use to minimize
costs?
Ans:
Let x denote the number of lb of premium loam used in making 100 lb of the
combination, and y the number of lb of generic loam. The Problem:
1
Minimize f ( x, y )  (5 x  y )
50
Subject to
0.6 x  0.2 y  36
i.e. 6 x  2 y  180
0.4 x  0.1y  20
4 x  1y  2 0 0
x  y  100
x  y 1 0 0
x, y  0 .
x, y  0 .
9. A crude insecticide commercially is 40% Toxin A and 35% Toxin B. New federal
regulations set upper limits on toxin levels for commercial insecticides: 36% for
Toxin A and 28% for Toxin B. A compatible insecticide can be produced using a
more refined process, but at an increased cost of $4 more than the crude insecticide
for every 10lb. This product would be only 15% Toxin A and 10% Toxin B. The
two insecticides can be blended. What combination of the two minimizes
production costs and meets federal standards?
Ans:
Let x denote the number of lb of the crude insecticide used in producing 100 lb of
an acceptable combination, and y the number of lb of the more refined insecticide.
The Problem:
1
Minimize f ( x, y )  (0 x  4 y )  0.4 y
10
Subject to
0.4 x  0.15 y  36
i.e. 8 x  3 y  720
0.35 x  0.1 y  28
7x  2 y  5 6 0
x  y  100
x  y 1 0 0
x, y  0 .
x, y  0 .
Section 2.3
3. The following are other suggested mathematical models for the problem of
Example lb (with the variables R, C, and X defined as in the example). Determine
why each model is not a proper representation of the problem.
(a) Maximize 50 R  60C  0.2(5 0 0)
Subject to
50R  30C  2 0 0 0
6R  5C  300
3R  5C  200
R, C  0 .
(b) Maximize
Subject to
50 R  60C  0.2(50 R  30C  1 5 0)0
50R  30C  2 0 0 0
6R  5C  300
3R  5C  200
R, C  0 .
(c) Maximize
Subject to
50R  60C  0.2 X
50R  30C  1500  X
6R  5C  300
3R  5C  200
X  500
R, C , X  0 .
Ans:
(a) Not true. Because it assume that 50R  30C is always equal to 2000:
50R  30C  2000 .
(b) Not true if 50R  30C  1500 .
(c) Not true. Because it assumed that 50R  30C is always at least 1500:
50R  30C  1500
8. A small plastic company makes novelty figures for sale at political conventions.
This spring the company has available 450 hr of labor and 825 spare units of
plastic for use in the production of donkeys and elephants. Each elephant requires
2 hr of labor and 7 units of plastic; each donkey requires 5 hr of labor and 5 units
of plastic. Elephants sell for $10, and up to 100 can be sold this summer; donkeys
sell for $7, with a market for 80. How many of each should the company produce
to maximize income over the coming summer?
Ans:
Let x(y) denote the member of elephants (donkeys) to be produced. The problem:
Maximize f ( x, y )  10 x  7 y
Subject to
2x  5 y  4 5 0
7 x  5 y  825
0  x  100
0  y  80 .
.
f (0,80)  f (25,80)  810
f (7 5,6 0)  1 1 7 0
f (1 0 ,00)  f (1 0 ,02 5)  1 1 7 5
Therefore max profit is $1175 attained by producing 100 elephants and 25
donkeys.
10. A cabinet shop produces and installs cabinets. Business is good, and the shop has
an unlimited number of customers willing to pay $100 for each cabinet installed.
However, for the next month, the shop has only 1750 man-hours of labor and 1032
units of wood that it can commit for cabinet production. Each installed cabinet
requires 5 hr of labor, 3 units of wood, and 1 frame. The frames can be prepared in
the shop before installation, with each frame requiring 2 hr of the shop’s labor and
1 unit of its wood; or, frames can be bought ready for instillation from the local
mill for $27 each. The shop pays $6/hr for labor, $5/unit for wood, and only pays
for the labor and wood used. For the next month, how many cabinets should the
shop install, and how should the necessary frames be generated, so that net income
is maximized?
Ans:
Let n denote the number of cabinets installed, and x(y) the number of frames
purchased (made internally). The problem:
Maximize 100n  27 x  6(5n  2 x)  5(3n  y )
Subject to
5n  2 y  1 7 5 0
3n  y  1 0 3 2
x y n
x, y , n  0 .
.
i.e.
Maximize f ( x, y )  28 x  38 y
Subject to
5n  2 y  1 7 5 0
3n  y  1032
x y n
x, y , n  0 .
.
The relevant vertices of (1) are (0,250), (224,90), (344,0); and
f (0,250)  9500
f (224,90)  9 6 9 2
f (3 4 ,40)  9 6 3 2
Therefore the maximum profit is $9692 attained by installing a total of 314 cabinets,
purchasing 224 of the necessary frames and making the remaining 90.
Section 2.4
Formulate mathematical models for the following problems. (Do not attempt to
solve the problems)
2. A canned goods supplier has two warehouses serving four outlets. The East Coast
Warehouse has 600 cases on hand and the West Coast Warehouse has 1000 cases
on hand. The shipping costs, in cents per case, and the requirements for the four
outlets, all located east of the Mississippi, are given in the following table.
Outlet 1
Outlet 2
Outlet 3
Outlet 4
East Coast Warehouse
20
16
30
20
West Coast Warehouse
45
39
50
44
300
350
400
450
Shipping costs
Requirements (cases)
Determine a shipping schedule that minimizes transportation costs.
Ans:
Let x1 j denote the number of cases shipped from the East Coast Warehouse to
Outlet j , j  1,2,3,4 , and let x 2 j denote the member of cases shipped from the
West Coast Warehouse to Outlet j , j  1,2,3,4 . Let x15 denote the number of
cases stored at the East Coast Warehouse and x 25 denote the number of cases
stored at the West Coast Warehouse. The problem:
Minimize 20 x11  16 x12  30 x13  20 x14  45x21  39 x22  50 x23  44 x24
subject to
5
x
j 1
ij
 600,1000, f o ri  1,2
ij
 300,350,400,450, f o rj  1,2,3,4
2
x
i 1
x ij is nonnegative integer, i  1,2; j  1,2,3,4,5 .
3. Three beverage plants supply five wholesale outlets with cases of soft drinks. The
weekly demands and transportation costs (in cents per case) are as follows.
Outlets
1
2
3
4
5
1
6.2
--
5.1
10.1
8.0
2
6.5
10.5
4.3
11.3
6.5
3
Requirements
(cases)
6.3
9.0
--
10.8
--
1000
1200
3000
400
2200
Plants
The dashes in the table indicate the impossibility of shipping cases between the
corresponding plants and outlets. The weekly production of Plant 1 is 4000 cases, of
Plant 2, 2000 cases, and of Plant 3, 3000 cases. Suppose also that the weekly surplus
at each plant can be sold locally for $1.20/case at Pant 1, $ 1.10/case at Plant 2, and
$1.14/case at Plant 3. Determine a shipping schedule that minimizes transportation
costs and that takes into account the amount accrued from the sale of the surplus.
Ans:
Introduce an additional outlet 6
demand: d 6 =(4000+2000+3000)-(1000+1200+3000+400+2200)=1200
‘cost’ for plant 1: c16  1.20
2: c26  1.10
3: c36  1.14
Let x ij denote the units shipped from Plant i to Outlet j, where i  1,2,3, j  1, ,6 .
The problem:
Minimize
 5.1x13  10.1x14  8 x15
6.2 x11
 120 x16
 6.5 x21  10.5 x22  4.3x23  11.3x24  6.5 x25  110 x26
 6.3x31  9.0 x32
 10.8 x34
 114 x36
Subject to
6
6
6
j 1
j 1
j 1
 x1 j  4 0 0,0 x2 j  2 0 0,0 x3 j  3 0 0 0
3
3
i 1
i 1
 xi1  1000,,  xi 6  1200
all xij  0, x12  x33  x35  0 .
4. A commodity is to be shipped from three warehouses to four outlets, each outlet
receiving 120 units. The shipping costs in dollars per unit are:
Outlets
1
2
3
4
1
12
15
10
25
2
10
19
11
30
3
21
30
18
40
Warehouses
Warehouse 1 has available 100 units, Warehouse 2, 150 units, and Warehouse 3, 300
units. For any unit not shipped, there is a storage charge of $6/unit at Warehouses 1
and 2 and $12/unit at Warehouse 3. Moreover, because of labor contracts, Outlet 2
cannot receive more units from Warehouse 1 than from Warehouse 2, and Outlet 4
must receive at least half of its supply from Warehouse 3. Determine a minimal cost
shipping and storing schedule.
Ans:
Let x ij denote the number of the units shipped from Warehouse i to Outlet j, and
xi 5 the number of units stored at warehouse i, where i  1,2,3, j  1,2,3,4 . The
problem:
Minimize
12 x11  15 x12  10 x13  25 x14  6 x15  10 x21  19 x22  11x23  30 x24  6 x25
 21x31  30 x32  18 x33  40 x34  12 x35
Subject to
5
x
j 1
ij
 1 0 ,01 5 ,03 0 0f o ri  1,2,3
3
x
i 1
ij
 1 2 ,0f o rj  1,2,3,4
x12  x22 , x34  60
xij  0, i  1,2,3; j  1,2,3,4 .
Section 2.5
Formulate mathematical models for the following problems. (Do not attempt to
solve the problems)
4. An appliance dealer sells small refrigerators to the college market. This July, 25
units are on hand. For the next 3 months, the dealer can buy from manufacture up
to 65 refrigerators each month, and can sell to the student population up to 100
units each month, at the following price.
Refrigerator
Buy($)
Sell($)
Aug.
60
90
Sept.
65
110
Oct.
68
105
The dealer has storage facilities for 45 units, but must pay a $7/unit/month storage
charge for each refrigerator stored for sale in a subsequent month. Determine an
optimal buying, selling and storing plan.
Ans:
For Aug., Sept., and Oct., let i=1,2,3
Bi  number of units of refrigerators bought
Ai  number of units of refrigerators sold
si  number of units of refrigerators stored
The problem:
Maximize 90 A1  110 A2  105 A3  (60B1  65B2  68B3 )  7( s1  s2  s3 )
Subject to
0  Ai  100, i  1,2,3
0  Bi  65, i  1,2,3
0  si  45, i  1,2,3
25  B1  A1  s1
s1  B2  A2  s2
s2  B3  A3  s3
8. A subsidiary division of an automobile plant produces automobile engines. For the
next four quarters, the demands of the plant are
Quarter
Number of Engines
1
2
3
4
400
450
800
550
There is an initial inventory of 100 engines. The division can produce 475 engines
in a quarter using its normal facilities. By the use of overtime, up to an additional
100 engines can be produced in any quarter, at a cost of $26/engine above the
normal costs. Any engines on hand at the end of a quarter can be stored at a cost of
$14/engine each quarter. Any quarterly demand not met by the division cost the
main plant in underutilization $33/engine for each quarter of the deficiency. By the
end of the fourth quarter, all the demands must be met. Determine an operating
schedule that minimizes costs.
Ans:
For Quarter i, let
Pi = number of engines produced, up to 475
Qi = number of engines over 475 produced
Di = number of engines delivered
Si = number of engines stored
U i = accumulated number of engines required but not delivered
The problem:
Minimize 26(Q1  Q2  Q3  Q4 )  14(S1  S 2  S3 )  33(U1  U 2  U 3 )
Subject to
0  Pi  475,
0  Qi  100
0  Di , Si , U i
100  P1  Q1  D1  S1
S1  P2  Q2  D2  S2
S2  P3  Q3  D3  S3
S3  P4  Q4  D4
D1 U1  400
D2  U 2  450  U1
D3  U 3  800  U 2
D4
 550  U 3