Linear Programming Problem

A dog food manufacturer makes 3 types of dog chew, each 10g in weight, which are made from different proportions of 2 basic ingredients. The table below shows this, together with the amount of the two ingredients in stock, and the costs for the three types of chew.

Chew type | Ingredient 1 | Ingredient 2 | Cost (p per chew)|

Type A                   8g                 2g                      1.8

Type B                   6g                 4g                      1.6

Type C                  5g                  5g                      1.5

Availability       800kg           400kg

The manufacturer wants to make 1600 packets of mixed chews. Each must contain 60 chews, and there must be no more than 30 of each type of chew in a packet.

a) Define appropriate variables and formulate an LP showing that:

8x+6y+5z<=500

2x+4y+5z<=250

 b) By eliminating the variable z show that the objective function can be given by  c = 90+0.3x+0.1y

and hence define the problem in terms of 2 variables.

c) By using a graphical approach solve the LP stating all the possible solutions.

d) From your graph you should see that 2 of the calculated constraints are unnecessary. Which constraints are these?

ANY HELP IS APPRECIATED!

Student X

Here's getting you started: 

Let x, y and z represent the number of each type of chews A, B and C respectively in a packet. 

Since 1600 packets are to be produced and a maximum of 800kg of ingredient 1 can be used in total, 800/1600= 0.5 kg or 500g is the maximum amount of ingredient 1 which can be used in a single packet.  

Similarly, since 1600 packets are to be produced and a maximum of 400kg of ingredient 2 can be used in total, 400/1600= 0.25kg or 250g is the maximum amount of ingredient 2 which can be used in a single packet. 

Hence, since the total amount of ingredients 1 and 2 used in making a single packet of chews are separately represented by 8x+6y+5z and 2x+4y+5z respectively, imposing the above conditions would yield 

8x+6y+5z<=500

2x+4y+5z<=250 (shown) 

We also know that since each packet must contact exactly 60 chews,

then  x+y+z=60 ====> z = 60-x-y---------(1) 

The cost involved in manufacturing one packet is c = 1.8x +1.6y +1.5z --------------(2) 

Substituting (1) into (2): 

c = 1.8x +1.6y +1.5(60-x -y) = 1.8x +1.6y + 90-1.5x -1.5y  

   = 90 +0.3x +0.1y (shown) 

Hopefully I have given you sufficient pointers to continue working on your own through the remaining parts of the question which should be pretty manageable. Peace.

Best Regards,

Mr Koh