1. 3.4 Linear Programming Objective:
Find the maximum and minimum values of a function over a region.

2. Linear Programming
Linear programming involves using a system of inequalities to find the maximum and minimum values to solve real-world problems.
Constraints – the actual lines of the inequalities
Feasible Region – the overlapping region of solutions
Vertices – the maximum or minimum of the feasible region.
Bounded – when the region is enclosed (a polygon)

3. Illustrations
Constraints
vertices
Feasible Region

4. Example
Graph the following system of inequalities. Name the coordinates of the vertices of the feasible region. Find the maximum and minimum values of the function
f(x, y)=3x-2y for this region.
x≤5
y≤4
x+y≥2 or y≥-x+2

5. Continued x≤5 y≤4 x+y≥2 or y≥-x+2 Find the vertices
y=-5+2 (substitute 5 for x)
y=-3 (5, -3)
4=-x+2 (substitute 4 for y)
x=-2 (-2, 4)
The first two lines intersect at (5, 4)
Use these vertices in the given function to find max and min. Maximum of 21 at (5,-3) Minimum of -14 at (-2,4)
(x, y)
3x-2y
f(x,y)
(5, -3)
3(5)-2(-3) 21
(-2, 4)
3(-2)-2(4)
-14
(5, 4)
3(5)-2(4) 7

6. Try One
y≥1 2≤x≤4 X-2y≥-4 f(x,y)=3y+x Solve for y (or x if there’s no y) 2≤x and x≤4 (change to x≥2) -2y≥-x-4 so y≤½x+2

7. Find vertices (x, y) 3y+x f(x,y) (2,1) 2(1)+2 4 (4,1) 3(1)+4 7 (2,3)
Some are easy (2,1) (4,1) Solve to find exact for other two: y≤½x+2 intersects x≥2 y=½(2) + 2 y=3 intersects at (2,3) y≤½x+2 intersects x≤4 y=½(4)+2 intersects at (4, 4)
Check the vertices with the given function Maximum: 16 at (4,4) Minimum: 4 at (2,1)
(x, y)
3y+x
f(x,y)
(2,1)
2(1)+2 4
(4,1)
3(1)+4 7
(2,3)
3(3)+2 11
(4,4)
3(4)+4 16

8. Unbounded
x≥-3 y≤1 3x+y≤6 (y≤-3x+6) f(x, y)=5x-2y Graph first Region is unbounded Vertex: (-3, 1) Vertex: 1=-3x+6 -5=-3x X=5/3 (5/3, 1)

9. Continued
Check to see about maximums and minimums. To determine if these are maximum/minimum points, check other points in the region. Choose (0,0) to see what happens in the function, f(x,y)=5(0)-2(0)= 0 This is larger than the minimum. Try (2,-3) f(x,y)=5(2)-2(-3)=16. This is larger than the above “maximum” in the table. That means that there is no maximum for this unbounded region. The minimum is at (-3,1)
(x,y)
5x-2y
f(x,y)
-3,1
5(-3)-2(1)
-17
5/3, 1
5(5/3)-2(1)
19/3

10. A landscaping company has crews who mow lawns and prune shrubbery
A landscaping company has crews who mow lawns and prune shrubbery. The company schedules 1 hour for mowing and hours for pruning jobs. Each crew is scheduled for no more than 2 pruning jobs per day. Each crew’s schedule is set up for a maximum of 9 hours per day. On the average, the charge for mowing a lawn is $40 and the charge for pruning a lawn is $120. Find a combination of mowing and pruning that will maximize the income the company receives per day for one of its crews.

12. Homework: Page odd