This process is covered in Section 4.4 of the textbook. For complete details on the process, consult this section and the many examples contained in this section. I will assume that you have looked over this section and are familiar with carrying out row operations on a matrix.

The problem we will solve is

Minimize *C* = 16*y*_{1} + 14*y*_{2} + 12*y*_{3 }subject to

*y*_{1} + *y*_{2} + *y*_{3} > 6000

2*y*_{2} – 3*y*_{3} > 0

3*y*_{1} – *y*_{2} – *y*_{3} > 0

*y*_{1} > 0, *y*_{2} > 0, *y*_{3} > 0

**Solution** Write out the dual problem and apply the Simplex Method. Start by entering the objective function and constraints into a matrix. The transpose of this matrix yields the form of the dual standard maximization problem.

The dual problem is

Maximize *z* = 6000*x*_{1} subject to

*x*_{1} + 3*x*_{3} < 16

*x*_{1} + 2*x*_{2} – *x*_{3} < 14

*x*_{1} – 3*x*_{2} – *x*_{3} < 12

*x*_{1} > 0, *x*_{2} > 0, *x*_{3} > 0

Now add the slack variables to each constraint and form the initial matrix for the Simplex Method.

The pivot column is the first column since that is where the indicator row is most negative. The pivot row is the third row. The entry in the pivot is already a zero so we simply need to put zeros in the rest of the column using row operations.

For this new matrix, the pivot is the 5 in the second row, second column. Change the pivot to a 1 and then use it to put zeros in the rest of the column. This results in

Since there is still a negative number in the indicator row, we continue applying the Simplex Method. In this step, the pivot is the 4 in the first row, third column. As before, this entry is changed to a 1 and used to make the rest of the column into zeros.

The indicator row does not contain any negatives numbers so this is the final matrix. The solution to the standard minimization problem is under the slack variables and indicate that

*y*_{1} = 1500, *y*_{2} = 2700, and *y*_{3} = 1800

with a corresponding minimum of *C* = 83400 cents or $834.

**Problem** – Garton’s Seeds has a seed mixture containing three types of seeds: bluegrass, rye, and Bermuda. The cost per pound of the three seeds are 16 cents, 14 cents and 12 cents. Bluegrass seed must be at least 25% of the each batch. The amount of Bermuda must be no more than 2/3 the amount of rye in each batch. To fill current orders, Garton’s must make at least 6000 lbs of the mixture. How many pounds of each seed should be in the batch so that the cost of the batch is minimized?

**Solution** Start by defining the variables to represent the seeds.

*y*_{1} – lbs of bluegrass in the mixture

*y*_{2} – lbs of rye in the mixture

*y*_{3} – lbs of Bermuda in the mixture

Since cost is to be minimized, let C be the cost of the mixture. We know the cost per pound on each type of seed. By multiplying the cost per pound times the number of pounds, we can find the cost of each type of seed. Adding the cost for each type gives

*C* = 16*y*_{1} + 14*y*_{2} + 12*y*_{3}

where *C* is in cents.

Now let’s look at the constraints. The easiest one to recognize relates to the requirement that at least 6000 lbs of the mixture must be made. This means that the sum of the variables representing the total amount of the mixture is greater than or equal to 6000,

*y*_{1} + *y*_{2} + *y*_{3} > 6000

In this format the inequality fits the form needed for a standard minimization. In this format the variables appear linearly on the left side of the inequality and are greater than or equal to a nonnegative number. Each of the inequalities must have this form to apply the Simplex Method.

Since the amount of Bermuda must be no more than 2/3 the amount of rye in each batch, we can write

*y*_{3} < 2/3 *y*_{2}

To put this in the proper format, subtract y_{3} from both sides and flip-flop the inequality to yield

2/3 *y*_{2} – *y*_{3} > 0

This has the proper format, but will be easier to deal with in the Simplex Method if there is no fraction. Multiply each term by 3 to give

2*y*_{2} – 3*y*_{3} > 0

The statement “Bluegrass seed must be at least 25% of the each batch” is a bit more difficult to interpret. On the surface, you might write

*y*_{1} > 0.25(6000)

thinking that 25% of the 6000 lb mixture must be bluegrass. But this would be incorrect since the batch will be at least 6000 lbs. Instead of 6000 lbs, write the total amount of the batch as *y*_{1} + *y*_{2} + *y*_{3}. Now we can interpret the information as

*y*_{1} > 0.25(*y*_{1} + *y*_{2} + *y*_{3})

Remove the parentheses and move all of the terms to the left side of the inequality. This yields

0.75*y*_{1} – 0.25*y*_{2} – 0.25*y*_{3} > 0

We can make this a bit more palatable by multiplying each term by 4,

3*y*_{1} – *y*_{2} – *y*_{3} > 0

Adding the nonnegativity constraints give the standard minimization problem:

Minimize *C* = 16*y*_{1} + 14*y*_{2} + 12*y*_{3 }subject to

*y*_{1} + *y*_{2} + *y*_{3} > 6000

2*y*_{2} – 3*y*_{3} > 0

3*y*_{1} – *y*_{2} – *y*_{3} > 0

*y*_{1} > 0, *y*_{2} > 0, *y*_{3} > 0

With the standard minimization problem established, we can now write out the dual problem and apply the Simplex Method.

]]>The basic algorithm for solving a standard maximization problem is covered in Section 4.3. This process, called the Simplex Method, uses matrices and row operations to gauge whether an objective function is maximized at corner points.

In the example below, I write out a standard maximization problem from an application and then solve it with the Simplex Method.

]]>Although a relative extrema may seem to be very similar to an absolute extrema, they are actually quite different. The term “relative” means compared to numbers nearby…so a relative extrema is either a bump or a dip on the function.

The term “absolute” means the most extreme on the entire function. An absolute extrema is the very highest or lowest point on the function. This may occur at a bump or a dip. They may also occur at the ends of the function if it is defined on a closed interval.

The MathFAQ below illustrates how to find these points on a function.

]]>Need help with applying the Union Rule for probability,

P(*E* or *F*) = P(*E*) + P(*F*) – P(*E* and *F*)

then the Math-FAQ “What Is The Rule For Computing The Probability Of A Union Of Two Events?” might be for you. Take a look and see how you can find probabilities involving “or” and “and”.

]]>In this Math-FAQ, learn how to compute factorials, permutations, and combinations using the Desmos scientific calculator.

]]>In a previous Math-FAQ, we looked at the different parts of a parabola. Based on this information, you know that to find the x intercepts of a parabola we need to solve a quadratic equation. When we solve a quadratic equation to find the x intercepts of the graph, you might expect to always have solutions. But as the Math-FAQ below shows, this is not always the case.

]]>and in another as

These are basically the same formulas, but used in a different context. However, how you solve for the different quantities in either one is the same. The Math-FAQs below demonstrate how to solve for

It can be difficult to distinguish between the bumps and dips on a graph, the relative (or local) extrema, and the very highest and lowest points on a graph, the absolute extrema.

The Math-FAQ belows show how to find each type of extrema using derivatives.

]]>Because of this, there are many FAQs available to help you work through these problems.

- Math-FAQ: How do you find the instantaneous rate of change?
- Math-FAQ: What is the difference between a secant line and a tangent line?
- Math-FAQ: How do you find the equation of a tangent line?
- Math-FAQ: How do you find a derivative at a point from the definition?
- Math-FAQ: How do you find the instantaneous rate from a table?

These examples should help you to solve problems from Section 11.2 and Section 11.3.

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