## How Do You Minimize Brewery Costs Under Some Constraints?

Let’s look at another example of writing out a standard minimization problem. This is a question about minimizing the costs at a brewery that produces regular and light beer. It goes something like this…

A brewery produces two types of beer. One beer is a regular beer with a typical number of calories. The other beer is a light beer containing a lower number of calories per serving. Steady customers of the brewery buy 10 units of regular beer and 15 units of light beer per month. The brewery wishes to produce extra beer beyond that which steady customers demand. The cost per unit of regular beer is 32,000 dollars and the cost per unit of light beer is 50,000 dollars. Every unit of regular beer brings in 120,000 dollars in revenue and every unit of light beer brings in 300,000 dollars in revenue. The brewery wants at least 9,000,000 dollars in revenue. At least 20 additional units of beer can be sold.

How much beer of each type must be sold to minimize production costs?

To begin a problem like this, define the independent variables:

R: units of regular beer

L: units of light beer

Once the variables are established, let’s write out the objective function that must be minimized. We are minimizing production costs and the pertinent info in the problem is

The cost per unit of regular beer is 32,000 dollars and the cost per unit of light beer is 50,000 dollars.

It is easiest to work in thousands of dollars and to write the objective function as

C = 32R + 50L

There are two constraints on production. First of all, the brewery needs at least 9,000,000 dollars (9000 thousand dollars). The corresponding revenue data in the problem is

Every unit of regular beer brings in 120,000 dollars in revenue and every unit of light beer brings in 300,000 dollars in revenue.

The revenue (in thousands of dollars from R units of regular beer is 120R and from L units of light beer is 300L. This means

120R + 300L ≥ 9000

The second constraint has to do with the fact that “at least 20 additional units of beer can be sold.” If steady customers require 10 units of regular beer and the brewery produces R units total, the additional amount produced is R – 10. Similarly, the additional amount of light beer produced is L – 15. So the constraint for additional units is

R – 10 + L – 15 ≥ 20

Isolating R and L on the left side gives

R + L ≥ 45

Let’s put all of this together in a standard minimization problem:

Minimize C = 32R + 50L subject to

120R + 300L ≥ 9000

R + L ≥ 45

R ≥ 0, L ≥ 0

The last constraints are need since the amounts of beer produced can’t be negative.

With this standard minimization problem in hand, we can now apply the simplex method to find the solution. Using this technique, we find that the minimum cost is 1800 thousand dollars (1,800,000 dollars) and occurs when 25 units of regular beer and 20 units of light beer are produced.

## Will a Graphical Solution Match the Simplex Solution?

Many standard maximization problems may be solved graphically or with the Simplex Method. If your problem has two independent variables (like x and y or x1 and x2),  you have a choice of which technique to utilize. However you do it, the solutions should be the same. Let’s look into this.

Maximize z = 5x1 + 2x2 subject to

2x1 + 4x2 < 15

3x1 + x2 < 10

x1 > 0, x2 > 0

Let’s look at the two solutions.

First off, here is the solution using a graphical approach.

In this solution, the feasible region is graphed using the inequalities. The corner points of the region are then tested in the objective function to find the maximum.

Now let’s look at the same problem solved with the Simplex Method.

First the problem needs to be written with slack variables and put into a matrix. Then the pivot is identified and row operations are used to change the pivot to a 1.

Once the pivot is a 1, the rest of the pivot column is filled with 0’s using row operations. Since the bottom row has negatives in it, another pivot is selected.

As before, the pivot is changed to a 1. Then the rest of the pivot column is changed to 0’s.

After these operations, there are no longer any negative numbers in the bottom row so we can read the solution off of the matrix. The solution, x1 = 2.5, x2 = 2.5 and z = 17.5, is the same as the graphical solution.

For problems with two independent variables, the linear programming problem can be found graphically and with the Simplex Method. For problems with more independent variables, the graphical method may not be applied so the Simplex Method becomes the best choice.