Linear Programming
Ex. A manufacturer of ice skates can produce as many as 60 pairs of hockey skates and as many as 45 pairs of figure skates per day. It takes 3 hours of labor to produce a pair of hockey skates and 4 hours of labor to produce a pair of figure skates. The company has up to 240 hours of labor available for ice skate production each day. If the profit is $12 on each pair of hockey skates and $18 on each pair of figure skates, find the number of pairs of each kind of skate the firm should produce to gain the maximum profit each day.
Solution:
Let x = hockey skates
y = figure skates
We begin by writing the equations/inequalities that we will use to determine the number of each kind of skate to produce. We can make this task easier by creating a table with the information.
|
Item produced |
Hours Required per skate |
Profit per skate |
|
Hockey skate (x) |
3 |
$12 |
|
Figure skate (y) |
4 |
$18 |
Available hours 240
Then
3x + 4y 240 The time to produce
hockey and figure skates cannot exceed 240 hours.
x 60
The firm can produce a maximum of 60 hockey skates in
a day.
y 45
The firm can produce a maximum of 45 figure skates in
a day.
12x + 18y The total profit the firm makes is $12 for each pair of hockey skates and $18 for each pair of
figure skates.
x 0
y 0
The amount produced must be a positive number.
This means when we graph the inequalities,
we will be limited to the 1st quadrant.
Now we graph the inequalities. since x and y must satisfy these inequalities, they are called constraints.
The shaded area is called the feasible region and the points of intersection are called corner points.
We now substitute the corner points into the profit equation to find where the maximum profit occurs.
(0,0) = 12(0) + 18(0) = 0
(60,0) = 12(60) = 18(0) = 720
(60,45) = 12(60) + 18(15) = 990
(20,45) = 12(20) + 18(45) = 1050
(0,45) = 12(0) + 18(45) = 810
Thus, the maximum profit is at (20,45). This means that the firm should produce 20 pairs of hockey skates and 45 pairs of figure skates to maximize their profit.