Question

Boys & Girls (a new e-dating service) undertakes two different online dating events that are popular items in the online-dating market: Hot (H) and Cool (C).  The profit margin on (Hot) H is $ 300; on C is $ 200.  For each couple that signs up with Hot (H), it takes the firm 6 hours of back-up checks (B), 4 hours of data entry (D), and 5 hours of interview (I). For each couple that signs up with Cool (C), it takes the 3 hours of B (back-up checks), 6 hours of D (data entry), and 5 hours of interview (I).  If 54 hours are available for B, 48 hours for D, and 50 hours for I, then how many slots of Hot (H) and Cool (C) should the company make available for potential customers.  Obviously, the company wants to maximize profits.              

State the objective function here:

Input/Output				Use this table to fill-in the details that describe the problem.  It will help in the next step.

Inequality form of the constraints	Equality form of the constraints (with slack)

Use this space to graph the constraints and derive the feasible set.  If you want you can scan the image and paste it here, or you can upload it in BB as another document.  Or the simplest is to copy and paste my figure from Unit 11 and relabel and change the entries on the axes.

Answer 1

Let the Hot(H) customers be represented by H and Cool(C) by C.

So the objective function is to maximize the profit i.e. 300*H + 200*C

The inequality functions would be, given the constraints because total time cannot exceed the constraint for each activity.

For B: 6H + 3C <=54

For D: 4H + 6C <=48

For I: 5H + 5C <=50

If we need to add the slack, then we need to put a variable s, wherein there is a gap between LHS and RHS.

Therefore, 6H + 3C + S1 = 54

4H + 6C + S2 = 48

5H + 5C + S3 = 50

where S1, S2, and S3 are the slack variables.

The picture of the graph has been attached. As it is clear from the graph, the feasible region is bounded by the points (0,0), (0,8), (9,0), and the intersection point of lines i.e. Point A(Intersection of lines I and D) and Point B(intersection of lines B and I).

Now, we will calculate these two points :

Point B (Lines I and B) : 6H + 3C =54 and 5H + 5C = 50

i.e. 30H + 15C = 270 and 30H + 30C = 300

Subtracting equations, we will get 15C = 30 i.e. C = 2

Putting value of C in any one equation, we get value of H to be 8. Therefore Point B is (8,2)

Point A (Lines D and I) : 5H + 5C = 50 and 4H + 6C =48

i.e. 20H + 20C = 200 and 20H + 30C = 240

Subtracting equations, we will get 10C = 40 i.e. C = 4

Putting value of C in any one equation, we get value of H to be 6. Therefore Point A is (6,4)

Now, we have the maximum value of the objective for each value of the bounded corner points

i.e. (0,0), (0,8), (9,0) , (8,2) and (6,4)

(0,0) : Value of objective function would be 0

(0,8) : Value of objective function would be 1600

(9,0) : Value of objective function would be 2700

(8,2) : Value of objective function would be 2800

(6,4) : Value of objective function would be 2600

We are getting the maximum value of the objective function at H = 8 and C =2.

Now, we will check the RHS and LHS of all constraints to check where we do have the constraint

For B: S1 = 54 - (6H + 3C)  = 0

For D: S2 = 48 - (4H + 6C) = 4

For I: S3 = 50 - (5H + 5C) = 0

As we can observe that we only have slack of 4 in equation of D. It means that we have 4 hours still left for data entry, even if we maximize our profits.