hours of finishing. The Company has 2 grinders and 3 finishers, each of whom work 40
hours per week.
Using-v to denote the number of dumpsters and у to denote the number of tankers,
write down a system of linear inequalities that describes the possible numbers of
each truck that can be manufactured.
(b) Graph the system and list its vertices.
(a)
24. Repeat Problem 23 if l grinder and 2 finishers, each of whom work 40 hours per week are
available.
3.3 A GEOMETRIC APPROACH TO LINEAR PROGRAMMING PROBLEMS
We begin by restating a portion of Example 9 given in the previous section: N u tt's
N uts has 75 pounds o f cashews and 120 pounds o f peanuts. These are to be mixed in l
pound packages as follows: a low-grade m ixture th at contains 4 ounces o f cashews
and 12 ounces o f peanuts and a high-grade m ixture that contains 8 ounces o f cashews
and 8 ounces o f peanuts.
Suppose that in addition to the inform ation given above, we also know w hat the
profit will be on each type o f mixture. For exam ple, suppose the profit is $0.25 on
each package o f the low-grade mixtures and is $0.45 on each package o f the highgrade mixture. The question o f im portance to the m anager is “ How m any packages
o f each type o f m ixture should be prepared to m aximize the profit?”
If P symbolizes the profit, .y is the num ber of packages of low-grade m ixture, and у
is the num ber o f high-grade packages, then the question can be restated as “ W hat are
the values o f л* and у so that the expression
P = $0.25* + $0.45 v
is a m axim um ?
Objective
Function
Constraints
The problem above is typical of a linear programming problem. It requires th at a
certain linear expression, the profit, be m aximized. This linear expression is called
the objective function. Furtherm ore, the problem requires that the m axim um profit
be achieved under certain restrictions or constraints, each o f which are linear inequal­
ities involving the variables. The linear program m ing problem may be restated as
M aximize
P = $0.25.V + $0.45 V
Objective function
subject to the conditions that
л >0
у 2=0
x + 2у 300
3.Y + 2y < 480
Nonnegativity constraint
Nonnegativity constraint
Cashew constraint
Peanut constraint
In general, every linear program m ing problem has two com ponents:
1. A linear objective function to be maximized or m inimized.
2. A collection o f linear inequalities that m ust be satisfied simultaneously.
L in ear P ro g ram m in g P ro b lem A linear programming problem in two variables, x
and y, consists of maximizing or minimizing an objective function
Z = A x + By
where A and В are given real num bers, subject to certain conditions or constraints
expressible as linear inequalities in x and y.
Feasible Solution
Let's look at this m ore closely. To maximize (or minimize) the quantity z =*
A y + By m eans to locate the points (.v, r) that make the expression for г the largest (or
smallest). But not all points (.v. r) are eligible. Only the points that obey all the
constraints are potential solutions. Hence, we refer to such points as feasible solu­
tions.
In a linear program m ing problem , we want to find the feasible solution that
maxim izes (or minimizes) the objective function.
Solution of a
Linear
Programming
Problem
By a solution to a linear programming problem we mean a point (x. y) in the set of
feasible solutions together with the value of the objective function at that point, that
maximizes (or minimizes) the objective function.
If none o f the feasible solutions maxim ize (or m inim ize) th e objective function, or
if there are no feasible solutions, then the linear program m ing problem has no
solution.
□ E xa m p le 1
M inim ize the quantity
г = л- + 21 subject to the constraints
V + Г S> 1
Solution
Д > ()
у
0
The objective function to be minimized is г = .v + 2y. The constraints are the linear
inequalities
X
+ г> I
л^0
Vs 0
The shaded portion of Figure 12 illustrates the set o f feasible solutions.
Figure 13
T o see if there is a smallest z, we graph z = x + 2у for some choice o f z, say. z = 3.
See Figure 14. By m oving the line x + 2 у = 3 parallel to itself, we can observe w hat
happens for different values o f z. Since we w ant a m inim um value for z, we try to
move z = x + 2у down as far as possible while keeping some part o f the line w ithin
the set of feasible solutions. The “ best” solution is obtained when the line ju st touches
one corner, or vertex, of.the set of feasible solutions. If you refer to Figure 14, you will
see that the best solution is x = l , у = 0, which yields z = l . There is no other feasible
solution for which z is smaller.
□
In Example l, we can see that the feasible solution that m inim izes z occurs at a
vertex. This is not an unusual situation. If there is a feasible solution m inim izing (or
maximizing) the objective function, it is usually located at a vertex o f th e set o f
feasible solutions.
However, it is possible for a feasible solution that is not a vertex to m inim ize (or
maximize) the objective function. This occurs when the slope o f the objective func­
tion is the same as the slope o f one side of the set o f feasible solutions. T he following
example illustrates this possibility.
□ E x am p le 2
M inimize the quantity
z = x + 2у
subject to the constraints
x + уs l
Solution
2x + 4 у > 3
x > 0
у> 0
Again, we first graph the constraints. The shaded portion of Figure 15 illustrates the
set o f feasible solutions.
If we graph the objective equation z = x + 2у for some choice o f z and move it
down, we see that a m inim um is reached when z = T In fact, any point on the line
2x + 4у = 3 between (^, £) and (4, 0) will m inim ize the objective function. O f course,
the reason any feasible point on
+ 4 y = 3 minimizes the objective equation
z = x + 2y is that these two lines are parallel (both have slope —4). Thus, this linear
program m ing problem has infinitely m any solutions.
□
The next example illustrates a linear program m ing problem that has no solution.
□ Exam ple 3
M aximize the quantity
z = x + 2y
subject to the constraints
X + У> l
Solution
Figure 16
X> 0
у 5: о
First, we graph the constraints. The shaded portion o f Figure 16 illustrates the set of
feasible solutions.
x+ у =
The graphs o f the objective function z = x + 2у for z = 2, z = 8, and z = 12 are
also shown in Figure 16. Observe that we continue to get larger values for z by moving
the graph o f the objective function upward. But there is no feasible point that will
m ake z largest. No m atter how large a value is assigned to z, there is a feasible point
that will give a larger value. Since there is no feasible point that makes z largest, we
conclude that this linear program m ing problem has no solution.
□
For any linear program m ing problem that has a solution, the following general
result is true:
If a linear program m ing problem has a solution, it is located at a vertex of the set
of feasible solutions; if a linear program m ing problem has multiple solutions, at
least one o f them is located at a vertex of the set of feasible solutions. In either
case, the corresponding value o f the objective function is unique.
The result stated above requires knowing in advance whether the linear program ­
ming problem has a solution. If the set o f feasible solutions is bounded — that is. if it
can be enclosed within some circle — the linear program m ing problem will have a
solution.
If the set o f feasible solutions is not bounded (see Example 3), then the graphs o f
the objective function for several values o f z should be used to determ ine w hether a
solution exists or does not exist.
Based on these com m ents, we can outline a procedure for solving a linear pro­
gram m ing problem provided that it has a solution.
Solving
a Linear
Programming
Problem
Step 1: W rite an expression for the quantity that is to be m axim ized or m ini­
mized (the Objective function).
Step 2: D eterm ine all the constraints and graph them.
Step 3: List the vertices o f the set o f feasible solutions.
Step 4: D eterm ine the value o f the objective function at each vertex.
Step 5: Select the optim al solution, that is, the m axim um or m inim um value o f
the objective function.
Let’s look at som e examples.
□ Exa m p le 4
M aximize and m inim ize the objective function
z = A' + 5y
subject to the constraints
©
Solution
л- -f 4 v < 12
©
x ^ 8
©
x + у
2
@ л- s 0
©
уs 0
The objective function and the constraints (num bered for convenience) are given
(this will not be the case when we do word problems), so we can proceed to graph the
constraints. The shaded portion o f Figure 17 illustrates the set o f feasible solutions.
Since this set is bounded, we know a solution exists.
Figure 17
Now we locate the vertices o f the set o f feasible solutions at the points o f intersec­
tion of lines © and @ , © and © , © and © , © and © , and © and @ . U sing
m ethods discussed earlier, we find that the vertices are
(0,3),
(8, 1),
(8, 0 ),
(2, 0 ),
(0 , 2)
To find the m axim um and m inim um value o f z = Л' + 5у, we set up a table:
Vertex
(*, j 0
Value of Objective Function
z = x + 5у
(0.3)
(8. 1)
(8. 0)
(2, 0)
(0. 2)
z=
z=
z=
z=
z=
0+
8T
8+
2+
0+
5(3) =
5(1)=
5(0) =
5(0) =
5(2) =
15
13
8
2
10
The m axim um value o f г is 15. and it occurs at the point (0. 3). The m inim um value
o f г is 2, and it occurs at the point (2, 0 ).
□
Now. let's solve the problem of the cashews and peanuts.
П E xam ple 5
Maximize
P = 0.25л- + 0.45 у
subject to the constraints
©
Solution
-v > 0
© у> 0
©
x + 2у < 300
@ 3.v + 2 v s; 480
Before applying the m ethod of this chapter to solve this problem, let's discuss a
solution that might be suggested by intuition. Namely, since the profit is higher for
the high-grade m ixture, you might think that N utt's Nuts should prepare as many
packages of the high-grade m ixture as possible. If this were done, then there would be
a total of 150 packages (8 ounces divides into 75 pounds o f cashews exactly 150 times)
and the total profit would be
150(0.45) = $67.50
As we shall see. this is not the best solution to the problem. This is because there
would be several pounds of peanuts left over (120 - 75 = 45, to be exact) that would
be neither packaged nor sold.
To use more o f the peanuts and thus make a higher profit. N utt's Nuts has to make
both high-grade and low-grade packages. We are still asking ourselves how many
packages o f each m ixture should be made to obtain the m axim um profit, but now we
will use linear program m ing to solve the problem. The graph of the set o f feasible
solutions is given in Figure 18.
Since this set is bounded, we proceed to locate its vertices. The vertices o f the set of
feasible solutions are the points o f intersection of lines (7) and (2), (7) and © , (2) and
@ , and © and @ :
(0 ,0 ),
(0,150),
(160,0),
(90,105)
(Notice that the points o f intersection of lines © and © and lines © and © are not
feasible solutions.) It rem ains only to evaluate the objective equation at each vertex:
Vertex
(a, y)
(0. 0)
(0, 150)
(160, 0)
(90, 105)
Value of Objective Function
P = ($0.25)* + ($0.45).)’
P=
P=
P=
P=
(0.25)(0) + (0.45)(0) = 0
(0.25)(0) + (0.45X150) = $67.50
(0.25)(160) + (0.45)(0) = $40.00
(0.25X90) + (0.45X105) = $69.75
Thus, a m axim um profit is obtained if 90 packages o f low-grade m ixture an d 105
packages of high-grade m ixture are made. The m axim um profit obtainable under the
conditions described is $69.75.
□
□ E xam ple 6
Solution
M ike’s Fam ous Toy Trucks m anufactures two kinds of toy tru ck s— a standard
model and a deluxe model. In the m anufacturing process, each standard m odel
requires 2 hours o f grinding and 2 hours of finishing, and each deluxe m odel needs 2
hours of grinding and 4 hours of finishing. The com pany has 2 grinders an d 3
finishers, each of whom work 40 hours per week. Each standard m odel toy truck
brings a profit of $3 and each deluxe model a profit o f $4. Assuming th at every truck
m ade will be sold, how m any o f each should be m ade to m axim ize profits?
First, we nam e the variables:
x = N um ber of standard models made
у = N um ber of deluxe models made
The quantity to be maximized is the profit, which we denote by P:
P = $3.y + $4 у
This is the objective function. To m anufacture one standard model requires 2 g rind­
ing hours and to make one deluxe model requires 2 grinding hours. Thus, the n u m ber
o f grinding hours of л* standard and у deluxe models is
2a + 2 у
But the total am ount of grinding tim e available is 80 hours per week. T his m eans we
have the constraint
2a + 2у ^ 80
Grinding time constraint
Similarly, for the finishing time we have the constraint
2a + 4 У ^ 120
Finishing time constraint
Simplifying each of these constraints and adding the nonnegativity constraints .v > 0
and у 2: 0 . we may list all the constraints for this problem.
л- + у < 40
.v + 2у ^ 60
.v ^ 0
y>0
Figure 19 illustrates the set o f feasible solutions, which is bounded.
Figure 19
The vertices of the set of feasible solutions are
(0 .0 ).
(0,30).
(40.0).
(20.20)
The table lists the corresponding values of the objective equation:
Vertex
(X,y)
(0. 0 )
(0, 30)
(40. 0)
(20. 20)
Value of Objective Function
P = S3x + $4y
P=0
P = %120
/>=$120
/>=3(20)+ 4(20) = $140
Thus, a m axim um profit is obtained if 20 standard trucks and 20 deluxe trucks are
m anufactured. The m axim um profit is $140.
□
□ E x a m pie 7
Inve stm e nt S trategy
A retired couple have up to $30.000 they wish to invest in fixed-income securities.
Their broker recom m ends investing in two bonds: one a AAA bond yielding 12%: the
other a B+ bond paying 15%. After some consideration, the couple decide to invest at
most $ 12.000 in the B erated bond and at least $6000 in the AAA bond. They also
want the am ount invested in the AAA bond to exceed or equal the am ount invested
in the B^ bond. W hat should the broker recom m end if the couple (quite naturally)
want to maximize their return on investment?
First, we nam e the variables:
-V= A m ount invested in AAA bond
У — A m ount invested in B+ bond
The quantity to be maximized — return on investm ent — which we denote by P. is
P = 0.12 a' + 0.15 j >
This is the objective function. The conditions specified by the problem are:
л* + у ^ 30.000
U p to $30,000 available to invest
Invest at most $ 12,000 in B+ bond
Invest at least $6000 in AAA bond
A m ount in AAA bond m ust exceed or equal am ount
in B+ bond
.
у ^ 12,000
x ^ 6.000
x 2: у
In addition, we m ust have the conditions.v ^ 0 and у ^ 0. The total list o f constraints
is
© x + у ^ 30.000
© j- < 12,000
® x s 6000
@
X
^
У
© A' £=0
© V> 0
Figure 20 illustrates the set of feasible solutions, which is bounded. The vertices o f the
set of feasible solutions are
(6000.0),
(6000,6000),
(12,000 12,000).
(18.000,12.000),
(30,000,0)
The corresponding return on investm ent at each vertex is:
P = 0.12(6000) + 0.15(0) = $720
P = 0.12(6000) + 0.15(6000) = 720 + 900 = $ 1620
P = 0.12(12,000) + 0.15(12,000) = 1 4 4 0 + 1800 = $3240
P = 0.12(18,000) + 0.15(12,000) = 2 1 6 0 + 1 8 0 0 = $3960
P = 0.12(30,000) + 0.15(0) = $3600
Thus, the m axim um return on investm ent is $3960, obtained by placing $ 18,000 in
the AAA bond and $12,000 in the B+ bond.
n
29.
30.
31.
32.
33.
34.
standard model is $4 and the profit on each deluxe model is $4. How many of each
should be manufactured in order to maximize profit?
Using the information supplied in Example 6 (page 13 1 ), suppose the profit on each
standard model is $4 and the profit on each deluxe model is $3. How many of each
should be manufactured in order to maximize profit?
Investment Strategy. An investment broker wants to invest up to $20,000. She can
purchase a type A bond yielding a 10% return on the amount invested and she can
purchase a type В bond yielding a 15% return on the amount invested. She also wants to
invest at least as much in the type A bond as in the type В bond. She will also invest at
least $5000 in the type A bond and no more than $8000 in the type В bond. How much
should she invest in each type of bond to maximize her return?
A factory manufactures two products, each requiring the use of three machines. The
first machine can be used at most 70 hours; the second machine at most 40 hours; and
the third machine at most 90 hours. The first product requires 2 hours on Machine 1,1
hour on Machine 2, and 1 hour on Machine 3; the second product requires 1 hour each
on Machines 1 and 2 and 3 hours on Machine 3. If the profit is $40 per unit for the first
product and $60 per unit for the second product, how many units of each product
should be manufactured to maximize profit?
Diet Problem. A diet is to contain at least 400 units of vitamins, 500 units of minerals,
and 1400 calories. Two foods are available: F ,, which costs $0.05 per unit, and ^ 2,
which costs $0.03 per unit. A unit of food F, contains 2 units of vitamins, 1 unit of
minerals, and 4 calories; a unit of food F2 contains 1 unit of vitamins, 2 units of
minerals, and 4 calories. Find the minimum cost for a diet that consists of a mixture of
these two foods and also meets the minimal nutrition requirements.
Diet Problem. Danny’s Chicken Farm is a producer of frying chickens. In order to
produce the best fryers possible, the regular chicken feed is supplemented by four
vitamins. The minimum amount of each vitamin required per 100 ounces of feed is:
Vitamin 1, 50 units; Vitamin 2, 100 units; Vitamin 3, 60 units; Vitamin 4, 180 units, j
Two supplements are available: Supplement I costs $0.03 per ounce and contains 5
units of Vitamin 1 per ounce, 25 units ofVitamin 2 per ounce, 10 units ofVitamin3 per
ounce, and 35 units ofVitamin 4 per ounce. Supplement II costs $0.04 per ounce and
contains 25 units ofVitamin I per ounce, 10 units ofVitamin 2 per ounce, 10 units of
Vitamin 3 per ounce, and 20 units ofVitamin 4 per ounce. How much of each supple- j
ment should Danny buy to add to each 100 ounces of feed in order to minimize his cost,
but still have the desired vitamin amounts present?
Optimal Use of Land. A farmer has 70 acres of land available on which to grow some
soybeans and some com. The cost of cultivation per-acre, the workdays needed per acre,
and the profit per acre are indicated in the table:
Soybeans
Corn
Total Available
$60
$30
$1800
Days of work per acre
3 days
4 days
120 days
Profit per acre
$300
Cultivation cost per acre
As indicated in the last column, the acreage to be cultivated is limited by the amount of
money available for cultivation costs and by the number of working days that can be put
into this part of the business. Find the number of acres of each crop that should be
planted in order to maximize the profit.
35. The manager of a supermarket meat department finds that there are 160 pounds of
round steak, 600 pounds of chuck steak, and 300 pounds of pork in stock on Saturday
morning. From experience, the manager knows that half these quantities can be sold as
straight cuts. The remaining meat will have to be ground into hamburger patties and
picnic patties for which there is a large weekend demand. Each pound of hamburger
patties contains 20% ground round and 60% ground chuck. Each pound of picnic
patties contains 30% ground pork and 50% ground chuck. The remainder of each
product consists of an inexpensive nonmeat filler which the store has in unlimited
quantities. How many pounds of each product should be made if the objective is to
maximize the amount of meat used to make the patties?
36. J. B. Rug Manufacturers has available 1200 square yards of wool and 1000 square yards
of nylon for the manufacture of two grades of carpeting: high-grade, which sells for $500
per roll, and low-grade, which sells for $300 per roll. Twenty square yards of wool and 40
square yards of nylon are used in a roll of high-grade carpet, and 40 square yards of
nylon are used in a roll of low-grade carpet. Forty work-hours are required to manufac­
ture each roll of the high-grade carpet, and 20 work-hours are required for each roll of
the low-grade carpet, at an average cost of $6.00 per work-hour. A maximum of 800
work-hours are available. The cost of wool is $5.00 per square yard and the cost of nylon
is $2.00 per square yard. How many rolls of each type of carpet should be manufactured
to maximize income? [Hint: Income = Revenue from sale —(Production cost for
material + labor)]
37. The rug manufacturer in Problem 36 finds that maximum income occurs when no
high-grade carpet is produced. If the price of the low-grade carpet is kept at $300 per roll,
in what price range should the high-grade carpet be sold so that income is maximized by
selling some rolls of each type carpet? Assume all other data remain the same.
38. Maximize
P = l x + у + 3z
subject to
x+2y + z^25
3 x + 2 y + 3 z = s3 0
*2>0
ys>0
2 2 0
[Hint: Solve the constraints three at a time, find the points that are in the set of feasible
solutions, and test each of them in the objective function. Assume a solution exists.]
CH AP TER R E V IE W
Important
Terms
linear inequality in two
variables
nonstrict linear inequality
strict linear inequality
graph of a linear inequality
half-plane
systems of linear inequalities
unbounded graph
bounded graph
vertex of the graph
linear programming problem
objective function
constraints
feasible solution
solution of a linear
programming problem