Optional: Minimisation problems

If you enjoyed using Geogebra to assist with decision-making, we have provided a couple of optional steps for you to take your learning a little further. (These optional steps will not be tested in the assessment.)

So far all of the problems we have considered have been maximisation problem, that is, we have been trying to maximise some quantity with respect to a set of constraints. However, some problems are minimisation problems rather than maximisation problems. They attempt to minimise some quantity with respect to some constraints.

The problem

A chemical manufacturing company generates three types of fertiliser – used to help plants grow – fertiliser A, fertiliser B and fertiliser C. The company has two factories which both manufacture all three types of fertiliser. Factory 1 can generate 100Kg of fertiliser A, 200Kg of fertiliser B and 300Kg of fertiliser C per day. It costs £10000 per day to operate the factory. Factory 2 can generate 200Kg of fertiliser A , 100Kg of fertiliser B and 200Kg of fertiliser C per day. It costs £9000 per day to operate the factory.

• Factory 1 is old and inefficient, for each day that it operates it produces 3000Kg of carbon dioxide. Factory 2 is slightly newer and only produces 2000Kg of carbon dioxide per day it operates.
• An order is placed for 2000Kg of fertiliser A and 2000Kg of fertiliser B and 3600Kg of fertiliser C.
• How many days should each refinery be operated for to fulfil the order at the smallest cost?
• How many days should each refinery be operated for to fulfil the order whilst minimising the amount of carbon dioxide produced?

We can solve this problem in a similar way as we did for maximisation problems. We start by formulating the problem:

Identify the decision variables

The decision variables are the number of days each of the factory is operated for. Let (x) be the number of days that factory 1 is operated for, and let (y) be the number of days that factory 2 is operated for.

Define the constraints

The amount of fertiliser A that is required for production is at least 2000Kg. Factory 1 can produce 100Kg per day and factory 2 can produce 200Kg per day.

(100x + 200y geq 2000)

The amount of fertiliser B that is required for production is at least 2000Kg. Factory 1 can produce 200Kg per day and factory 2 can produce 100Kg per day.

(200x + 100y geq 2000)

The amount of fertiliser C that is required for production is at least 3600Kg. Factory 1 can produce 300Kg per day and factory 2 can produce 200Kg per day.

(300x + 200y geq 3600)

Finally, it is not possible to operate a chemical factory for less than zero days so we also have the constraints,

(x geq 0), and

(y geq 0)

Objective function

The cost of operating the chemical factories is the summation of the cost of operating factory 1 multiplied by the number of day and the cost of operating factory 2 multiplied by the number of days. The cost per day to operate factory 1 is £10000, and the cost to operate factory 2 for a pay in £9000.

(C = 10000x+9000y)

Formulation

Minimise (C = 10000x + 9000y)

Subject to (begin{align*}  100x + 200y &leq 2000 \ 200x + 100y &leq 2000 \ 300x + 200y &leq 3600\ x &geq 0\ y &geq 0. end{align*})

Using the graphical method

We can plot the constraints in a graph (below). This time the feasible region is unbounded as any point in the unshaded region is feasible. . However, this isn’t a problem as we are trying to minimise the cost. When we try to minimise the objective function we first select a point in the feasible region. When we were maximising we were attempting to move the line representing the objective function away from the origin. When we want to minimise, we want to move the line representing the objective function towards the origin. The last point in the feasible region as we move the objective function line towards the origin is the optimal solution.

We can find the vertices that lower bound the feasible by solving the following simultaneous equations.

The vertex ((0, 20)) is the intersection of the lines (x=0) and (200x +100y =2000). By substituting (x=0) into (200x +100y = 2000) we get (y=20) The cost of not using factory 1, and running factory 2 for 20 days is (10000 times 0 +9000 times 20 = 180 000), that is £180 000.

The vertex ((4,12)) is the intersection of the lines (200x +100y = 2000) and (300x + 200y =3600). The cost of running factory 1 for 4 days, and running factory 2 for 12 days is (10000 times 4 +9000 times 12 = 148000), that is £148 000.

The vertex ((8,6)) is the intersection of the lines (300x +200y=3600) and (100x +200y=2000). The cost of running factory 1 for 8 days, and running factory 2 for 6 days is (10000 times 8 +9000 times 6 = 134000), that is £134 000.

The vertex ((20,0)) is the intersection of the lines (100x+200y=2000)and (y=0). By substituting (y=0) into (100x+200y=2000) we get (x=20.)The cost of running factory 1 for 20 days, and running factory 2 for 0 days is (10000 times 20 +9000 times 0 = 200 000), that is £200 000.

The minimum cost is realised when factor 1 is operated for 8 days and factory 2 is operated for 6 days. 

Using Geogebra

If you would like to use Geogebra to do this work, below is a Geogebra file containing the constraints.