1. Today’s Agenda
 Attendance
/ Announcements
 Questions from Yesterday
 Sections 7.2
 Quiz Today

2. Exam Schedule
 Exam
4 (Ch 6,7)
Fri 11/15
 Exam 5 (Ch 10)
Thur 12/5
 Final Exam (All)
Thur 12/12

3. Linear Programming
Businesses use linear
programming to find out how to
maximize profit or minimize
costs. Most have constraints on
what they can use or buy.

4. Linear Programming
The Objective Function is
what we need to maximize or
minimize. For us, this will be a
function of 2 variables, f(x, y)

5. Linear Programming
The Constraints are the
inequalities that provide us with
the Feasible Region.

6. Linear Programming (pg 400)

7. The general idea… (pg 398)
Find max/min values of the objective
function, subject to the constraints.
Objective Function
f ( x, y)
2x 5 y
Constraints
3x 2 y 6
2x 4 y 8
x y 1
x 0, y 0

8. The general idea… (pg 398)
Graph the Feasible Region

9. The general idea… (pg 398)
The Feasible Region makes up the possible
inputs to the Objective Function
f ( x, y)
2x 5 y

10. Corner Point Thm (pg 400)
If a feasible region is
bounded, then the
objective function
has both a
maximum and
minimum value,
with each occurring
at one or more
corner points.

11. Find the minimum and maximum
value of the function f(x, y) = 3x - 2y.
We are given the constraints:
• y≥2
• 1 ≤ x ≤5
• y≤x+3

12. 1 ≤ x ≤5
8
7
6
5
4
3
y≤x+3
y≥2
2
1
1
2
3
4
5

13. Need to find corner
points (vertices)
1 ≤ x ≤5
8
7
6
5
4
3
y≤x+3
y≥2
2
1
1
2
3
4
5

14. • The vertices (corners) of the
feasible region are:
(1, 2) (1, 4) (5, 2) (5, 8)
• Plug these points into the
function f(x, y) = 3x - 2y
Note: plug in BOTH x, and y values.

15. Evaluate the function at each vertex
to find min/max values
f(x, y) = 3x - 2y
• f(1, 2) = 3(1) - 2(2) = 3 - 4 = -1
• f(1, 4) = 3(1) - 2(4) = 3 - 8 = -5
• f(5, 2) = 3(5) - 2(2) = 15 - 4 = 11
• f(5, 8) = 3(5) - 2(8) = 15 - 16 = -1

16. So, the optimized solution is:
• f(1, 4) = -5 minimum
• f(5, 2) = 11 maximum

17. Find the minimum and maximum value
of the function f(x, y) = 4x + 3y
With the constraints:
y
y
y
x 2
1
x 2
4
2x 5

18. y ≥ 2x -5
Need to find corner
points (vertices)
6
5
y ≤ 1/4x + 2
4
y ≥ -x + 2
3
2
1
1
2
3
4
5

19. Evaluate the function at each vertex
to find min/max values
f(x, y) = 4x + 3y
• f(0, 2) = 4(0) + 3(2) = 6
• f(4, 3) = 4(4) + 3(3) = 25
7
1
1
7
• f( 3 , - 3 ) = 4( 3 ) + 3(- 3 ) =
28
3
-1 =
25
3

20. So, the optimized solution is:
• f(0, 2) = 6 minimum
• f(4, 3) = 25 maximum

21. Classwork / Homework
• Page 403
• 1, 3, 7, 9, 11