3.4 Linear Programming DAY 2 Objectives:   To find maximum and minimum values

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3.4 Day 2 Linear Programming 2010
October 27, 2010
3.4 Linear Programming
DAY 2
Objectives: • To find maximum and minimum values
• To solve problems with linear programming
Aug 14­9:04 PM
1
3.4 Day 2 Linear Programming 2010
October 27, 2010
Check Skills You'll Need:
Solve each system of inequalities by graphing.
1. x > 5
y > ­3x + 6
2. 3y > 5x + 2
y < ­x + 7
3. x + 3y < ­6
2x ­ 3y < 4
Aug 14­9:31 PM
2
3.4 Day 2 Linear Programming 2010
October 27, 2010
Solving Real­World Problems:
Suppose you are selling cases of mixed nuts and roasted peanuts. You can order no more than a total of 500 cans and packages and spend no more than $600. How can you maximize your profit? How much is the maximum profit?
Mixed Nuts
Roasted Peanuts
12 cans per case
20 packages per case
You pay...$24 per case
Sell at...$3.50 per can
You pay...$15 per case
Sell at...$1.50 per package
$18 profit per case!
$15 profit per case!
Aug 14­9:44 PM
3
3.4 Day 2 Linear Programming 2010
October 27, 2010
Solving Real­World Problems:
Suppose you are selling cases of mixed nuts and roasted peanuts. You can order no more than a total of 500 cans and packages and spend no more than $600. How can you maximize your profit? How much is the maximum profit?
Define:
Let x = number of cases of mixed nuts ordered
Let y = number of cases of roasted peanuts ordered
Let P = total profit
Relate:
Organize the information into a table
Number of Cases
Number of Units
Cost
Profit
Mixed Nuts
x
12x
24x
18x
Roasted Peanuts
y
20y
15y
15y
Total
x + y
500
600
18x + 15y
constraint
constraint
objective
Sep 28­4:50 PM
4
3.4 Day 2 Linear Programming 2010
October 27, 2010
Solving Real­World Problems:
Suppose you are selling cases of mixed nuts and roasted peanuts. You can order no more than a total of 500 cans and packages and spend no more than $600. How can you maximize your profit? How much is the maximum profit?
Number of Cases
Number of Units
Cost
Profit
Mixed Nuts
x
12x
24x
18x
Roasted Peanuts
y
20y
15y
15y
Total
x + y
500
600
18x + 15y
constraint
constraint
objective
Write: Write and simplify the constraints. Write the objective function.
{
12x + 20y < 500
24x + 15y < 600
x > 0, y > 0
⇒
{
3x + 5y < 125
8x + 5y < 200
x > 0, y > 0
P = 18x + 15y
Now follow the steps from yesterday to determine what values of x and y maximize your profit.
Sep 28­4:50 PM
5
3.4 Day 2 Linear Programming 2010
October 27, 2010
Step 1: Graph the constraints (solve for y first)
{
3x + 5y < 125
8x + 5y < 200
x > 0, y > 0
⇒
{
y < ­3/5x + 25
y < ­8/5x + 40
x > 0, y > 0
50
40
30
Step 2: Find the coordinates
of each vertex
20
10
10
20
30
40
50
(0, 0)
(25, 0)
(15, 16)
(0, 25)
Sep 28­4:50 PM
6
3.4 Day 2 Linear Programming 2010
October 27, 2010
Step 3: Evaluate P at each vertex
P = 18x + 15y
(0, 0)
(25, 0)
(15, 16)
(0, 25)
P = 18(0) + 15(0) = 0
P = 18(25) + 15(0) = 450
P = 18(15) + 15(16) = 510
P = 18(0) + 15(25) = 375
Step 4: State the results in complete sentences.
You can maximize your profit by selling 15 cases of mixed nuts and 16 cases of roasted peanuts. The maximum profit is $510.
Sep 28­4:50 PM
7
3.4 Day 2 Linear Programming 2010
October 27, 2010
Homework: page 142 (10, 11, 20, 21, 23 ­ 27)
Sep 28­5:38 PM
8
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