# SECTION 3.4 �Linear Programming�

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```					    SECTION 3.4
“Linear Programming”
WHAT’S IMPORTANT:
-- Solve linear programming problems.
-- Be able to use linear programming to
solve real-life problems.
Section 3.4: “Linear Programming”
 Optimization: finding the maximum or
minimum value of some quantity
 Linear Programming: the process of
optimizing a linear “objective function”
subject to a system of linear inequalities
called “constraints”
 Feasible Region: the graph of the system
of constraints
Section 3.4: “Linear Programming”
   If an objective function has a maximum of a
minimum value, then it must occur at a vertex of the
feasible region. Moreover, the objective function will
have both a maximum and a minimum value if the
feasible region is “bounded”.

Bounded Region             Unbounded Region
Section 3.4: “Linear Programming”
 Example: Find the minimum and
maximum values of C = -x + 3y subject to
the following constraints:      x2
x5
y0
y  -2x + 12
 Begin by graphing the lines for the
constraints. (All will be solid lines.)
Section 3.4: “Linear Programming”
   Each mark is 2 units.
   Graph x ≥ 2
   Graph x ≤ 5
   Graph y ≥ 0
   Graph y ≤ -2x + 12
Section 3.4: “Linear Programming”
 Fill in the vertices and shade in the region.
 As you can see, the vertices are at (2,0),
(5,0), (5,2), and (2,8).
Section 3.4: “Linear Programming”
 Since this is a bounded region, there will
be both a minimum and a maximum.
 Use the points of the vertices in the
equation C = -x + 3y to find them.
Section 3.4: “Linear Programming”
   Put in (2,0): C = -x + 3y = -2 + 3(0)
= -2 + 0 = -2
   Put in (5,0): C = -x + 3y = -5 + 3(0)
= -5 + 0 = -5
   Put in (5,2): C = -x + 3y = -5 + 3(2)
= -5 + 6 = 1
   Put in (2,8): C = -x + 3y = -2 + 3(8)
= -2 + 24 = 22
   So, the minimum is -5 and the maximum is 22.
Section 3.4: “Linear Programming”
   Example: Find the minimum and
maximum values of C = x + 5y subject to
the following constraints: x0
y  2x + 2
5x+y

   Begin by graphing the lines for the
constraints. (All will be solid lines.)
Section 3.4: “Linear Programming”
 Each mark is 1 unit.
 Graph x ≥ 0
 Graph y ≤ 2x + 2
 Graph 5 ≥ x + y
Section 3.4: “Linear Programming”
 Fill in the vertices and shade the
appropriate region.
 The vertices are at (0,0), (0,2), (1,4), and
(5,0)
Section 3.4: “Linear Programming”
 Since this is an unbounded region, there
will be a maximum, but not a minimum.
 Use the points of the vertices in the
equation C = x + 5y to find the maximum.
Section 3.4: “Linear Programming”
 Put in (0,0): C = x + 5y = 0 + 5(0)
=0+0=0
 Put in (0,2): C = x + 5y = 0 + 5(2)
= 0 + 10 = 10
 Put in (1,4): C = x + 5y = 1 + 5(4)
= 1 + 20 = 21
 Put in (5,0): C = x + 5y = 5 + 5(0)
=5+0=5
 So, the maximum is 21. No minimum.
Section 3.4: “Linear Programming”
   Example: A furniture manufacturer makes
chairs and sofas. The table on the next slide
gives the number of packages of wood, stuffing,
and fabric required for each chair or sofa. The
manufacturer has room for 1300 packages of
wood parts, 2000 packages of stuffing, and 800
packages of fabric. The manufacturer earns
\$200 per chair and \$350 per sofa. How many
chairs and sofas should they try to make to
maximize profit?
Section 3.4: “Linear Programming”
MATERIAL           CHAIR              SOFA
WOOD              2 Boxes           3 Boxes
STUFFING           4 Boxes           3 Boxes
FABRIC             1 Box            2 Boxes

Using this and the info in the problem, we can figure
the profit formula and the constraint formulas.
Letting x = the # of chairs sold and y = # of sofas
sold, P = \$200(x) + \$350(y)
Section 3.4: “Linear Programming”
MATERIAL          CHAIR            SOFA
WOOD             2 Boxes         3 Boxes
STUFFING          4 Boxes         3 Boxes
FABRIC            1 Box          2 Boxes

He can only store 1300 packages of wood, so one
constraint is: 2x + 3y  1300
Likewise, 4x + 3y  2000 and x + 2y  800
And, since he cannot produce a negative number of
items, x  0 and y  0
Section 3.4: “Linear Programming”
   Each mark is 100 units.
   Graph 2x + 3y ≤ 1300
   Graph 4x + 3y ≤ 2000
   Graph x + 2y ≤ 800
   And graph x ≥ 0 and y ≥ 0
Section 3.4: “Linear Programming”
 Fill in the vertices and shade the
appropriate region.
 The vertices are at (0,0), (0,400),
(200,300), (300,200), and (500,0)
Section 3.4: “Linear Programming”
   Use the vertices in the equation
P = \$200x + \$350y to find the maximum.
Section 3.4: “Linear Programming”
   At (0,0):     P = 200(0) + 350(0)
= 0 + (0) = \$0
   At (0,400):   P = 200(0) + 350(400)
= 0 + 1400 = \$1400
   At (200,300): P = 200(200) + 350(300)
= 40,000 + 105,000 = \$145,000
   At (300,200): P = 200(300) + 350(200)
= 60,000 + 70,000 = \$130,000
   At (500,0):   P = 200(500) + 350(0)
= 100,000 + 0 = \$100,000
   So, the maximum profit is \$145,000 at (200,300).
   The manufacturer needs to sell 200 chairs and
300 sofas to maximize profits.

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