LINEAR PROGRAMMING
WORD PROBLEMS WITH ANSWERS
1. Mary works selling cards over the telephone. She sells two types of
cards, birthday cards and holiday cards. Mary makes $2.00 for each box of
birthday cards she sells, and she makes $2.50 for each box of holiday cards
she sells. She can work no more than 10 hours per week. It takes her an
average of 15 minutes to sell one box of birthday cards and an average of
20 minutes to sell one box of holiday cards. If she can sell no more than 35
boxes of cards in total, how many boxes of each type should she sell to
make the most money? (Hint: remember that an hour contains 60 minutes!)
x – birthday cards
y – holiday cards
x 0
y 0
P = 2x + 2.50y
15 x  20 y  600
y 
3
x  30
4
x  y  35
y   x  35
Vertices:
(0, 0) = $0.00
(0, 30) = $75.00
(20, 15) = $77.50
(35, 0) = $70.00
Answer: 20 birthday cards and 15 holiday cards; max profit = $77.50
2. Your factory makes fruit filled breakfast bars and granola bars. For each
case of breakfast bars, you make a $40 profit. For each case of granola bars,
you make a $55 profit. It takes 2 machine hours to make a case of breakfast
bars and 5 hours of labor. It takes 6 machine hours and 4 labor hours to
make s case of granola bars.You have a maximum of 150 machine hours
and 160 labor hours available. How many cases of each product should you
produce in order to maximize profit?
x – breakfast bar
y – granola bar
x 0
y 0
2 x  6 y  150
y 
1
x  25
3
5 x  4 y  160
5
y   x  40
4
P = 40x + 55y
Vertices:
(0, 0) = $0.00
(0, 25) = $1375.00
(15,20) = $1700.00
(33,0) = $1320.00
Answer: approximately 15 cases of breakfast bars and 20 cases of granola bars;
max profit - $1700.00
3. An oil refinery has a maximum production of 2,000 barrels of oil per
day. It produces two types of petroleum products; gasoline and heating oil.
The government requires that the refinery produce at least 300 barrels of
heating oil per day. If the profit is $3.00 per barrel of gasoline and $2 a
barrel for heating oil, how much of each petroleum product should the oil
refinery produce per day to maximize profit? What is the maximum profit
per day?
x – gasoline
y – heating
P = 3x + 2y
x 0
y  300
x  y  2000
Vertices:
(0, 300)
= $600.00
y   x  2000 (0, 2000) = $4000.00
(1700, 300)= $5700.00
Answer: 1700 barrels of gasoline and 300 barrels of heating;
max profit - $5700.00
4. A Virginia Beach farmer has 480 acres of land on which to grow either
corn or soybeans. He figures he has 800 hours of labor available during the
crucial summer season. The farmer can expect a profit of $40 per acre on
corn and $30 per acre on soybeans. He knows that corn requires 2 hours
per acre to raise, and soybeans require 1 hour per acre to raise. How many
acres of each should he plant to maximize profit? What is his maximum
profit?
x – acre of corn
y – acre of soybeans
x 0
y 0
x  y  480
y   x  480
P = 40x + 30y
Vertices:
(0, 0) = $0.00
(0, 480) = $14400.00
2 x  y  800
(400, 0) = $16000.00
y  2 x  800 (320, 160) =$17600.00
Answer: 320 acres of corn and 160 acres of soybeans; max profit - $17,600.00
5. Brad owns a news stand that has room for 100 newspapers per day. In
his town, there are two papers, The Journal and The Globe. Every day, Brad
sells 20 Journals and 25 Globes to customers with subscriptions. If Brad
makes $0.05 for every Journal sold and $0.10 for each Globe sold, how
many Journals and how many Globes should he put on his stand to make
the most money?
x – # of The Journal
y – # of The Globe
x  20
y  25
x  y  100
y   x  100
P = .05x + .10y
Vertices:
(20, 25) = $3.50
(20, 80) = $9.00
(75, 25) = $6.25
Answer: 20 The Journal and 80 The Globe; max profit - $9.00
6.Your club plans to raise money by selling two sizes of fruit baskets. The
Plan is to buy small baskets for $10 and sell them for $16, then buy large
baskets for $15 and sell them for $25. The club president estimates that
you will not sell more than 100 baskets.Your club can afford to spend up to
$1,200 to but baskets. Find the number of small and large fruit baskets you
should buy in order to maximize profit.
x – # of small fruit baskets
y – # of large fruit baskets
x
y
xy
y
0
0
 100
  x  100
10 x  15 y  1200
2
y   x  80
3
P = 6x + 10y
Vertices:
(0, 0) = $0.00
(0, 80) = $800.00
(100, 0) = $600.00
(60, 40) = $760.00
Answer: 0 small fruit baskets and 80 large fruit baskets; max profit - $800.00
LINEAR PROGRAMMING
ACTIVITY SHEET
WORD PROBLEMS WITH ANSWERS
1. The Buds-R-Us Florist has to order roses and carnations for Valentine’s
Day. Roses cost the florist $20 per dozen and carnation cost $5 per
dozen. The profit on roses is $20 per dozen and on carnations it is $8 per
dozen. The florist can spend at most $450 on flowers and must order at
least 20 dozen carnations, but the florist can order no more than 60 dozen
flowers. How many of each kind should he order in order to maximize the
profit?
x – dozens of roses
y – dozens of carnations
y  20
x  y  60
y   x  60
20 x  5 y  450
y  4 x  90
P = 20x + 8y
Vertices:
(10, 50) = $600.00
(~17, 20)= $500.00
(40, 20) = $960.00
Answer: 40 dozens of roses and 20 dozens of carnations; max profit - $960.00
2. U. N. Scrupulous, a noted businessman in town, always uses both
newspaper and radio advertising each month. It is estimated that each
newspaper ad reaches 8,000 people and that each radio ad reaches 15,000
people. Each newspaper ad costs $50 and each radio ad costs $100. The
business can spend no more than $1000 for advertising per month. The
newspaper requires at least 4 ads be purchased each month and the radio
requires that at least 5 ads be purchased each month. What combination of
newspaper and radio ads should old “Scrupe” purchase in order to reach
the maximum number of people?
x – newspaper ad
y – radio ad
x4
y 5
50 x  100 y  1000
1
y   x  10
2
P = 8000x + 15000y
Vertices:
(4, 8) =152,000 people
(4, 5) =107,000 people
(10, 5) =155,000 people
Answer: 10 newspaper ads and 5 radio ads; max people reached – 155,000
3. The campus store sells stadium cushions and caps. The cushions cost
$1.90; the caps cost $2.25. They sell the cushions for $5.00 and the caps
for $6.00. They can obtain no more than 100 cushions and 75 caps per
week. To meet the demands, they have to sell a total of at least 120 of the
two items together. They cannot package more than 150 per week. How
many of each should they sell to maximize profit?
x – cushions
y – caps
x  100
P = 3.1x + 3.75y
y  75
x  y  120
y   x  120
x  y  150
y   x  150
Vertices:
(45, 75) = $420.75
(75, 75) = $513.75
(100, 20)= $385.00
(100, 50) =$497.50
Answer: 75 cushions and 75 caps; max profit - $513.75