Introduce variables and write inequalities for the conditions given in the question.
Skim: 179 gallons Whole: 21 gallons
Practice makes perfect
We can use linear programming to find the amount of skim and whole milk the Yodel Family Dairy should produce to maximize their profit.
Setting up the Inequalities
First, let's assign variables to each type of milk.
Description
Variable
Amount of skim milk in gallons.
s
Amount of whole milk in gallons.
w
We can use these variables to express the conditions given in the exercise.
Description
Inequality
The Yodel Family Dairy produces at most 200 gallons of milk.
s+w≤ 200
Regular customers need at least 15 gallons of skim milk.
15≤ s
Regular customers need at least 21 gallons of whole milk.
21≤ w
We can write our boundaries as a system of inequalities.
s+w≤ 200 & (I) 15 ≤ s & (II) 21 ≤ w & (III)
Graphing the Inequalities
We can graph these inequalities to get an illustration of the constraints on the daily milk production.
Graph of Inequality I
The graph of s+w≤ 200 is a half plane bounded by the line s+w=200. We can graph this line through its intercepts, (200,0) and (0,200).
To complete the graph, we will test a point that is not on the line and decide which region we should shade. Let's test point ( 0, 0). If the point satisfies the inequality, we will shade the region that contains the point. Otherwise, we will shade the other region.
Since this inequality is true, we shade the region that contains the test point (0,0).
Graph of Inequality II
The inequality 15≤ s describes all values of s that are greater than or equal to 15. This means that every coordinate pair with a s-value that is greater than or equal to 15 will be included in the shaded region.
Graph of Inequality III
The inequality 21≤ w describes all values of w that are greater than or equal to 21. This means that every coordinate pair with a w-value that is greater than or equal to 21 will be included in the shaded region.
Combining the Inequality Graphs
Finally, we will draw the graphs of the inequalities on the same coordinate plane.
Now, we really only need to consider the overlapping region. Any point (s,w) in this region (or on the boundary) represents a possible arrangement of a daily production of skim and whole milk.
Setting up the Objective Function
Let's turn our attention now to the goal. We need to optimize the profit.
It is given that the profit on a gallon of skim milk is $0.82, and the profit on a gallon of whole milk is $0.75.
Using these values, we can express the daily profit (P) if the dairy produces s gallons of skim milk and w gallons of whole milk.
P=0.82s+0.75w
Finding the Maximum Profit
The maximum profit will correspond to one of the vertices of the feasible region we found above. Let's highlight these vertices on the diagram.
We will need the coordinates of the vertices.
We can find these as the intersection points of the corresponding lines.
Point
Lines
Coordinates
A
15=s 21=w
ls=15 w=21
B
15=s s+w=200
ls=15 w=200-15=185
C
s+w=200 21=w
ls=200-21=179 w=21
We can substitute these coordinates in the profit equation and see which point gives the maximum profit.
Let's start with A(15,21).
We can find the profit corresponding to the other vertices in a similar way.
Vertex
P=0.82s+0.75w
Value
A( 15, 21)
0.82( 15)+0.75( 21)
28.05
B( 15, 185)
0.82( 15)+0.75( 185)
151.05
C( 179, 21)
0.82( 15)+0.75( 21)
162.53
We can see that the production of s=179 gallons of skim milk and w=21 gallons of whole milk gives the maximum daily profit of $162.53 for the Yoder Family Dairy.
Alternative Solution
Alternative way of thinking
This question can also be solved without linear programming by noticing several things about the problem and putting them in the correct order.
The profit on a gallon of skim milk is greater than the profit on a gallon of whole milk. This means that if the Yodel Family Dairy has a choice, they should choose to produce skim milk over whole milk.
They need to produce at least 21 gallons of whole milk. According to the first observation, it is best for them to stop there and use the rest of their capacity to produce skim milk.
The more they produce, the greater the profit is. The maximum amount they can produce is 200 gallons. According to the second observation, 21 gallon of this should be whole milk and the rest, 200-21=179 gallons should be skim milk.
Using this information, we can conclude that the Yodel Family Dairy can maximize their daily profit by producing 179 gallons of skim milk and 21 gallons of whole milk.
