Transcript finding a maximum profit model

10TH
EDITION
COLLEGE
ALGEBRA
LIAL
HORNSBY
SCHNEIDER
5.6 - 1
5.6
Systems of Inequalities and Linear
Programming
Solving Linear Inequalities
Solving Systems of Inequalities
Linear Programming
5.6 - 2
Solving Linear Inequalities
A linear inequality in two variables is an
inequality of the form
Ax  By  C,
where A, B, and C are real numbers, with A
and B not both equal to 0. (The symbol 
could be replaced with , <, or >.) The graph
of a linear inequality is a half-plane, perhaps
with its boundary.
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Solving Linear Inequalities
For example, to graph the linear inequality
3 x  2y  6,
first graph the boundary.
Since the points of
the line satisfy the
equation, this line is
part of the solution
set.
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Solving Linear Inequalities
To decide which half-plane (the one above the
line 3x – 2y = 6 or the one below the line) is part of
the solution set, solve the original inequality for y.
3 x  2y  6
2y  3 x  6
Reverse the
inequality symbol
when dividing by a
negative number.
3
y  x 3
2
Subtract 3x.
Divide by − 2.
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Solving Linear Inequalities
For a particular value
of x, the inequality will
be satisfied by all
values of y that are
greater than or equal
to 32 x  3. Thus, the
solution set contains
the half-plane above
the line.
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Caution A linear inequality must be in
slope-intercept form (solved for y) to
determine, from the presence of a < symbol or
a > symbol, whether to shade the lower or
upper half-plane. In the previous slide, the upper
half-plane is shaded, even though the inequality
is 3x – 2y  6 is (with a < symbol) in standard
3
y

x 3
form. Only when we write the inequality as
2
(slope-intercept form) does the > symbol indicate
to shade the upper half-plane.
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Example 1
GRAPHING A LINEAR
INEQUALITY
Graph x + 4y > 4.
Solution
The boundary of the graph is the straight
line x + 4y = 4. Since points on this line do
not satisfy x + 4y > 4, it is customary to
make the line dashed. To decide which
half-plane represents the solution set, solve
for y.
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Example 1
GRAPHING A LINEAR
INEQUALITY
x  4y  4
4y   x  4
1
y   x 1
4
Subtract x.
Divide by 4.
1
 x  1,
4
Since y is greater than
the graph of the
solution set is the half-plane above the boundary.
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GRAPHING A LINEAR
Example 1
INEQUALITY
Alternatively, or as a check, choose a test point not on
the boundary line and substitute into the inequality.
The point (0, 0) is a good choice if it does not lie
on the boundary, since the substitution is easily done.
x  4y  4
0  4(0)  4
04
Original inequality.
Use (0, 0) as a
test point.
False
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Example 1
GRAPHING A LINEAR
INEQUALITY
Since the point (0, 0) is below the boundary, the
points that satisfy the inequality must be above the
boundary, which agrees with the result.
xy 4
0  4(0)  4
04
Original inequality.
Use (0, 0) as a
test point.
False
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Graphing Inequalities
1. For a function , the graph of y < (x) consists of all the
points that are below the graph of y = (x); the graph of
y > (x) consists of all the points that are above the
graph of y = (x).
2. If the inequality is not or cannot be solved for y, choose
a test point not on the boundary. If the test point
satisfies the inequality, the graph includes all points on
the same side of the boundary as the test point.
Otherwise, the graph includes all points on the other
side of the boundary.
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Solving Systems of Inequalities
The solution set of a system of inequalities, such
as
x  6  2y
x  2y ,
2
is the intersection of the solution sets of its
members. We find this intersection by graphing the
solution sets of all inequalities on the same
coordinate axes and identifying, by shading, the
region common to all graphs.
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Example 2
GRAPHING SYSTEMS OF
INEQUALITIES
Graph the solution set of each system.
a.
x  6  2y
x 2  2y
Solution
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GRAPHING SYSTEMS OF
Example 2
INEQUALITIES
Graph the solution set of each system.
b. x  3
y 0
y  x 1
Solution
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Example 2
GRAPHING SYSTEMS OF
INEQUALITIES
Writing x  3 as –3  x  3 shows that this
inequality is satisfied by points in the region
between and including x = –3 and x = 3.
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Example 2
GRAPHING SYSTEMS OF
INEQUALITIES
The set of points that satisfies y  0 includes
the points below or on the x-axis.
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Example 2
GRAPHING SYSTEMS OF
INEQUALITIES
Graph y = x + 1 and use a test point to
verify that the solutions of y  x + 1 are on
or above the boundary.
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Example 2
GRAPHING SYSTEMS OF
INEQUALITIES
Since the solution sets of y  0 and y  x +1
have no points in common, the solution set
of the system is ø.
The solution set of the system is ø, because
there are no points common to all three regions.
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Note
While we gave three graphs in
the solutions of Example 2, in practice
we usually give only a final graph showing
the solution set of the system.
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Solving a Linear
Programming Problem
Step 1 Write the objective function and all
necessary constraints.
Step 2 Graph the region of feasible solutions.
Step 3 Identify all vertices or corner points.
Step 4 Find the value of the objective function at
each vertex.
Step 5 The solution is given by the vertex
producing the optimal value of the
objective function.
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FINDING A MAXIMUM PROFIT
MODEL
The Charlson Company makes two products—
MP3 players and DVD players. Each MP3 gives a
profit of $30, while each DVD player produces $70
profit. The company must manufacture at least 10
MP3s per day to satisfy one of its customers, but
no more than 50 because of production problems.
The number of DVD players produced cannot
exceed 60 per day, and the number of MP3s
cannot exceed the number of DVD players. How
many of each should the company manufacture to
obtain maximum profit?
Example 3
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Example 3
FINDING A MAXIMUM PROFIT
MODEL
Solution
First we translate the statement of the problem into
symbols.
Let x = number of MP3s to be produced daily,
and y = number of DVD players to be produced daily.
The company must produce at least 10 MP3s (10 or
more), so
x  10.
Since no more than 50 MP3s may be produced,
x  50.
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Example 3
FINDING A MAXIMUM PROFIT
MODEL
Solution
No more than 60 DVD players may be made in
one day, so
y  60.
The number of MP3s may not exceed the number
of DVD players translates as
x  y.
The numbers of MP3s and of DVD players cannot
be negative, so
x0
and
y  0.
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FINDING A MAXIMUM PROFIT
MODEL
Example 3
Solution
These restrictions, or constraints, form the system
of inequalities
x  10,
x  50,
y  60,
x  y,
x  0,
y  0.
Each MP3 gives a profit of $30, so the daily profit
from production of x MP3s is 30x dollars. Also, the
profit from production of y DVD players will be
70y dollars per day. Total daily profit is, thus,
profit  30 x  70y .
This equation defines the function to be maximized,
called the objective function.
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Example 3
FINDING A MAXIMUM PROFIT
MODEL
Solution
To find the maximum
possible profit, subject to
these constraints, we
sketch the graph of each
constraint.
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Example 3
FINDING A MAXIMUM PROFIT
MODEL
Solution
The only feasible values
of x and y are those that
satisfy all constraints—
that is, the values that lie
in the intersection of the
graphs of the constraints.
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Example 3
FINDING A MAXIMUM PROFIT
MODEL
Solution
Any point lying inside the
shaded region or on the
boundary satisfies the
restrictions as to the
number of MP3s and
DVD players that may be
produced.
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Example 3
Solution
FINDING A MAXIMUM PROFIT
MODEL
(For practical purposes,
however, only points with
integer coefficients are
useful.) This region is called
the region of feasible
solutions. The vertices
(singular vertex) or corner
points of the region of feasible
solutions have coordinates
(10,10), (10,60), (50,50),
and (50,60).
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FINDING A MAXIMUM PROFIT
MODEL
We must find the value of the objective function
30x + 70y for each vertex. We want the vertex that
produces the maximum possible value of 30x + 70y.
Example 3
(10,10) : 30(10)  70(10)  1000
(10,60) : 30(10)  70(60)  4500
(50,50) : 30(50)  70(50)  5000
(50,60) : 30(50)  70(60)  5700
Maximum
The maximum profit, obtained when 50 MP3s and
60 DVD players are produced each day, will be
30(50) + 70(60) = 5700 dollars per day.
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Example 3
FINDING A MAXIMUM PROFIT
MODEL
To show why the point of the feasible solution works:
The Charlson Company needed to find values
of x and y in the shaded region that produce
the maximum profit—that is, the maximum
value of 30x + 70y. To locate the point (x, y)
that gives the maximum profit, add to the graph
lines corresponding to arbitrarily chosen profits
of $0, $1000, $3000, and $7000:
30 x  70 y  0, 30 x  70 y  1000,
30 x  70 y  3000, and 30 x  70 y  7000.
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Example 3
FINDING A MAXIMUM PROFIT
MODEL
Each point on the line 30x + 70y = 3000 corresponds
to production values that yield a profit of $3000.
The region of feasible
solutions are shown with
these lines. The lines are
parallel, and the higher the
line, the greater the profit.
The line 30x + 70y = 7000
yields the greatest profit but
does not contain any points
of the region of feasible
solutions.
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Example 3
FINDING A MAXIMUM PROFIT
MODEL
To find the feasible
solution of greatest profit,
lower the line
30x + 70y = 7000 until it
contains a feasible
solution- that is, until it just
touches the region of
feasible solutions. This
occurs at point A, a vertex
of the region.
The result observed here hold for every linear
programming problem.
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Fundamental Theorem of
Linear Programming
If an optimal value for a linear programming
problem exists, it occurs at a vertex of the
region of feasible solutions.
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Example 4
FINDING A MINIMUM COST
MODEL
Robin takes vitamin pills each day. She
wants at least 16 units of Vitamin A, at least
5 units of Vitamin B1 and at least 20 units of
Vitamin C. She can choose between red
pills, costing 10 cents each, that contain 8
units of A, 1 of B1 and 2 of C; and blue pills,
costing 20 cents each, that contain 2 units of
A, 1 of B1 and 7 of C. How many of each pill
should she buy to minimize her cost and yet
fulfill her daily requirements?
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Example 4
FINDING A MINIMUM COST
MODEL
Since Robin buys x of the 10 cent pills and
y of the 20 cent pills, she gets 8 units of
Vitamin A from each red pill and 2 units of
Vitamin A from each blue pill. Altogether
she gets 8x +y units if A per day. Since she
wants at least 16 units,
8 x  2y  16.
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Example 4
FINDING A MINIMUM COST
MODEL
Each red pill and each blue pill supplies 1 unit of
Vitamin B1. Robin wants at least 5 units per day,
so
x  y  5.
For Vitamin C, the inequality is
2 x  7 y  20.
Also, x  0 and y  0, since Robin cannot buy
negative numbers of the pills.
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Example 4
FINDING A MINIMUM COST
MODEL
Step 2 The intersection of the graphs of
8 x  2y  16,
x  y  5,
x  0,
and
2 x  7 y  20,
y 0
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Example 4
FINDING A MINIMUM COST
MODEL
Step 3 The vertices are (0, 8), (1, 4), (3, 2), and
(10, 0).
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FINDING A MINIMUM COST
MODEL
Steps 4 and 5 We find that the minimum cost
occurs at (3, 2).
Point
Cost = 10x + 20y
Example 4
(0, 8)
10(0) + 20(8) = 160
(1, 4)
10(1) + 204) = 90
(3, 2)
1(3) + 20(2) = 70
(10, 0)
10(10) + 20(0) = 100
Minimum
Robin’s best choice is to buy 3 red pills and 2
blue pills, for a total cost of 70 cents per day. She
receives just the minimum amounts of Vitamins
B1 and C, and an excess of Vitamin A.
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