Transcript 3.4 Linear Programming

3.4 Linear Programming
Steps
• 1. Solve for y and graph each inequality.
• 2. Find the common area that the inequalities
cover. Determine the boundaries of this common
area.
• 3. Find locations where the graph changes
direction. Mark these points. These are your
constraint points.
• 4. You will be given an equation. To determine
which constraint point maximizes the equation,
plug each constraint point into the equation
What is Linear Programming?
Linear Programming is a technique that
identifies the minimum and maximum
value of some quantity. The quantity is
modeled with an objective function, a
separate equation that is used to
determine what the maximum and
minimum constraint points of the region
will be.
What is a constraint?
A constraint is a limit on the variables in
the objective function. The constraints are
the linear inequalities that are used to find
the feasible region.
Let’s revisit the two graphs
we looked at last lecture.
GRAPHING CONSTRAINTS
x + y ≥ -1
x–y≤6
y≤2
y ≥ -x – 1
y≥x–6
y≤2
(8 , 2)
(-3 , 2)
Which point maximizes f(x,y) = x + 2y ?
(-3
, 2)
f(-3,2) = -3 + 2(2) = 1
_____________________________
(8
, 2)
f(8,2) = 8 + 2(2) = 12
_____________________________
(2.5 , 3.5) f(2.5,3.5) = 2.5 + 2(3.5) = 9.5
_____________________________
(2.5 , 3.5)
GRAPHING CONSTRAINTS
2 ≤ x ≤ 6 2 ≤ x and x ≤ 6
1 ≤ y ≤ 5 1 ≤ y and y ≤ 5
x + y ≤ 8 y ≤ -x + 8
(3, 5)
(6, 2)
(2, 5)
Which point maximizes f(x,y) = x – 2y ?
(2,
5)
f(2,5) = 2 – 2(5) = -8
_____________________________
(3 , 5)
f(3,5) = 3 – 2(5) = -7
_____________________________
(6 , 2)
f(6,2) = 6 – 2(2) = 2
_____________________________
(6 , 1)
f(6,1) = 6 – 2(1) = 4
_____________________________
(2 , 1)
f(2,1) = 2 – 2(1) = 0
_____________________________
(2, 1)
(6, 1)
Using a graphing calculator, let’s graph
the following inequalities with an
objective function of P = 13x + 2y
-3x + 2y < 8
-8x + y >-48
x>0
y>0
First step: get
the equations
into slopeintercept form
(solve for y)
y < 3/2x +4
y >8x -48
x>0
y>0
• Now let’s use our Graphing Calculator.
Take out your Linear Programming Cheat
Sheet.
Linear Programming
Example #1
• You want to buy tickets
for some concerts; rock and jazz.
• You want at least 5 rock tickets
and at least 8 jazz tickets.
• No more than 25 tickets total.
• Rock tickets cost $30 and
jazz tickets cost $20.
• You cannot spend more than $600. Find the
max # of tix each you can buy at the lowest
price total.
STEP 1
Set up Equations
Use each piece of information given to
make an equation or restraint.
For our example
x = rock Y= jazz
At least 5 rock tix  x > 5
At least 8 jazz tix  y > 8
No more than 25 tickets  x + y < 25
Rock tix costs $30 , jazz tix cost $20. No more than $600
 20x + 30y < 600
Objective function 
P=20x+30y
STEP 2
GRAPH
Use your calculator
x>5
y>8
x + y < 25
20x + 30y < 600
STEP 3
POINTS OF INTERSECTION
X = rock
Y = jazz
(5,16.6)
(15,10)
(17,8)
(5,8)
STEP 4
TEST VERTICES INTO OBJECTIVE FUNCTION
Now it is time to test the vertices to see what
combination of x and y, rock and jazz respectively,
will give us the max number of tickets at the lowest
price.
Objective Notation
(x , y)
(5 , 8)
(15 , 10)
(17 , 8)
(5 , 16.6)
f(x , y) = 20x + 30y
f(5, 8) = 20(5) + 30(8)
= $ 340
f(15, 10) = 20(15) + 30(10)
= $ 600
f(17, 8) = 20(17) + 30(8)
= $ 580
f(5, 16.6) = 20(5) + 30(16.6)
= who cares..?
Linear Programming
Example #2
• A pizza shop makes $1.50 on each small
pizza and $2.15 on each large pizza.
• On a typical day, it sells between 70 and
90 small pizzas and between 100 and 140
large pizzas.
• The shop can make no more than 210
pizzas in a day.
• How many of each size of pizza must be
sold to maximize profit?
STEP 1
Set up Equations
Use each piece of information given to
make an equation or restraint.
For our example
x = small pizzas
Y= large pizzas
Between 70 and 90 small pizzas 70 < x < 90
Between 100 and 140 large pizzas  100 < y < 140
Shop can make no more than 210 pizzas x + y < 210
Objective function
A pizza shop makes $1.50 on each small pizza
and $2.15 on each large pizza.

P=1.50x+2.15y
STEP 2
GRAPH
Use your calculator
70 < x < 90
100 < y < 140
x + y < 210
140
130
120
110
100
70
80
90
100
STEP 3
POINTS OF
INTERSECTION
(70,140)
(90,120)
140
130
120
(70,100)
110
100
70
80
90
100
(90,100)
STEP 4
TEST VERTICES INTO OBJECTIVE FUNCTION
Now it is time to test the vertices to see what
combination of x and y, small and large pizzas
respectively, will give us the max profit.
(x , y)
Objective Notation
f(x , y) = 1.50x + 2.15y
(70 , 140) f(70, 140) = 1.50(70) + 2.15(140)
= $ 406
(90 , 120) f(90, 120) = 1.50(90) + 2.15(120)
= $ 393
(70 , 100) f(70, 100) = 1.50(70) + 2.15(100)
= $ 320
(90 , 100) f(90, 100) = 1.50(90) + 2.15(100)
= $ 350