Maximize R = 90x + 120y Simplex Method Explained

Hey guys! Today, we're diving into a classic optimization problem using the simplex method. It might sound intimidating, but trust me, we'll break it down step by step. Our goal is to maximize the objective function R = 90x + 120y, but we've got some constraints to work with. Think of it like trying to make the most profit while dealing with limited resources. These constraints are given by the following inequalities:

$$\begin{array}{l} 3 x+2 y \leq 24 \\ x+2 y \leq 20 \\ x \geq 0, y \geq 0 \end{array}$$

These constraints define a feasible region – basically, the set of all possible (x, y) values that satisfy our limitations. And remember, x and y have to be non-negative, which makes sense in many real-world scenarios (you can't produce a negative amount of something!). Let's get started on how to solve this problem using the simplex method!

Setting Up the Initial Simplex Tableau

The first thing we need to do in the simplex method is transform our inequalities into equations. We do this by introducing slack variables. These variables, which we'll call 's' and 't', represent the unused resources in each constraint. This is a crucial step because the simplex method operates on systems of equations, not inequalities. So, our inequalities become:

$$\begin{aligned} 3x + 2y + s &= 24 \\ x + 2y + t &= 20 \end{aligned}$$

Here, 's' represents the slack in the first constraint (3x + 2y ≤ 24), and 't' represents the slack in the second constraint (x + 2y ≤ 20). Now, we also need to rewrite our objective function (R = 90x + 120y) to include it in the tableau. We move all the terms to the left side of the equation, setting it equal to zero:

$$-90x - 120y + R = 0$$

Now we have a system of equations that we can neatly organize into a simplex tableau. The tableau is a matrix that represents our system, making it easier to perform the necessary row operations. Our initial simplex tableau looks like this:

	x	y	s	t	R	RHS
s	3	2	1	0	0	24
t	1	2	0	1	0	20
R	-90	-120	0	0	1	0

Let's break down what each part represents:

• The top row represents our variables: x, y, s, t, and R (our objective function). The last column, RHS, stands for "Right Hand Side" and contains the constant values.
• Each row represents one of our equations. The first two rows correspond to our constraints (including slack variables), and the last row represents our objective function.
• The entries in the tableau are the coefficients of the variables in our equations. For example, in the first row, 3, 2, 1, 0, and 0 are the coefficients of x, y, s, t, and R, respectively, in the equation 3x + 2y + s = 24.

This initial tableau represents our starting point. We've successfully converted our problem into a format that the simplex method can work with. Now, we're ready to start iterating towards the optimal solution! In the next step, we'll identify the pivot column, which will guide us in improving our objective function's value. Understanding the setup of this initial tableau is absolutely crucial, so make sure you've got it down before moving on. It's the foundation for everything else we'll do!

Identifying the Pivot Column and Pivot Row

Alright, now that we have our initial simplex tableau set up, the next crucial step is to figure out how to improve our solution. Remember, we're trying to maximize R, so we want to find a way to increase its value. This is where the concepts of the pivot column and pivot row come into play. These elements will guide us through the process of pivoting, which is how we move from one feasible solution to a better one.

Pivot Column

The pivot column is the column that corresponds to the variable that will enter our solution basis. Think of it as the variable that, if we increase its value, will most significantly increase the value of our objective function (R). To find the pivot column, we look at the last row of our tableau – the row representing our objective function. Specifically, we look for the most negative entry in this row.

Why the most negative? Because these negative entries represent the amount by which R will increase for each unit increase in the corresponding variable. The more negative the entry, the greater the potential increase in R. In our tableau:

	x	y	s	t	R	RHS
s	3	2	1	0	0	24
t	1	2	0	1	0	20
R	-90	-120	0	0	1	0

We see that -120 is the most negative entry. This corresponds to the 'y' column, so the y column is our pivot column. This tells us that increasing the value of 'y' will have the biggest positive impact on R.

Pivot Row

Now that we've identified the pivot column, we need to determine the pivot row. The pivot row tells us which variable will leave our solution basis. This is a bit trickier than finding the pivot column, as it involves a ratio test. We perform the ratio test to ensure that we don't violate any of our constraints while increasing the value of our pivot column variable ('y' in this case).

Here's how the ratio test works:

1. Divide the RHS (Right Hand Side) values by the corresponding entries in the pivot column. We only consider rows where the entry in the pivot column is positive (if it's zero or negative, it doesn't make sense for our ratio calculation).
2. Identify the smallest non-negative ratio. The row corresponding to this smallest ratio is our pivot row.

Let's apply this to our tableau. Our pivot column is 'y', so we divide the RHS values (24 and 20) by the corresponding entries in the 'y' column (2 and 2):

• For the 's' row: 24 / 2 = 12
• For the 't' row: 20 / 2 = 10

The smallest ratio is 10, which corresponds to the 't' row. Therefore, the t row is our pivot row. This means that the variable 't' will leave our basis as we increase the value of 'y'.

Pivot Element

Finally, the pivot element is the entry at the intersection of the pivot column and the pivot row. In our case, this is the entry in the 't' row and 'y' column, which is 2. The pivot element is crucial because we'll use it to perform row operations and transform our tableau in the next step.

So, to recap:

• Pivot Column: 'y' (because -120 is the most negative entry in the last row)
• Pivot Row: 't' (because 10 is the smallest non-negative ratio)
• Pivot Element: 2

With the pivot column, pivot row, and pivot element identified, we're ready to perform the pivoting operation. This is where the real magic happens, and we'll see how our tableau transforms to give us a better solution. Stick with me, guys, we're making great progress!

Pivoting: Transforming the Tableau

Okay, we've reached the heart of the simplex method: pivoting. Pivoting is the process of systematically transforming our tableau to improve our solution. It involves performing row operations to make the pivot element equal to 1 and all other entries in the pivot column equal to 0. This process essentially swaps the roles of the entering variable (from the pivot column) and the leaving variable (from the pivot row) in our solution basis.

Remember, our current tableau looks like this, with the pivot element (2) highlighted:

	x	y	s	t	R	RHS
s	3	2	1	0	0	24
t	1	2	0	1	0	20
R	-90	-120	0	0	1	0

Step 1: Make the Pivot Element Equal to 1

To do this, we divide the entire pivot row (the 't' row) by the pivot element (2). This gives us:

New 't' row = (Old 't' row) / 2

	x	y	s	t	R	RHS
s	3	2	1	0	0	24
t	1/2	1	0	1/2	0	10
R	-90	-120	0	0	1	0

Notice that the 'y' entry in the 't' row is now 1, as desired. This is a key step because it normalizes the pivot row, making it easier to eliminate the other entries in the pivot column.

Step 2: Make All Other Entries in the Pivot Column Equal to 0

Now, we need to eliminate the 'y' entries in the other rows (the 's' row and the 'R' row). We do this by using row operations that add or subtract multiples of the new pivot row (the transformed 't' row) from the other rows.

• Eliminate 'y' from the 's' row:

  • We want to make the 'y' entry in the 's' row (which is currently 2) equal to 0. To do this, we subtract 2 times the new 't' row from the 's' row:

  New 's' row = (Old 's' row) - 2 * (New 't' row)

  	x	y	s	t	R	RHS
  s	2	0	1	-1	0	4
  t	1/2	1	0	1/2	0	10
  R	-90	-120	0	0	1	0

• Eliminate 'y' from the 'R' row:

  • We want to make the 'y' entry in the 'R' row (which is currently -120) equal to 0. To do this, we add 120 times the new 't' row to the 'R' row:

  New 'R' row = (Old 'R' row) + 120 * (New 't' row)

  	x	y	s	t	R	RHS
  s	2	0	1	-1	0	4
  t	1/2	1	0	1/2	0	10
  R	-30	0	0	60	1	1200

And there you have it! We've completed one full pivot. Our tableau has been transformed, and the 'y' variable has entered the basis, replacing 't'. Notice that the value of R has increased from 0 to 1200. This indicates that we've moved closer to the optimal solution.

Interpreting the New Tableau

Before we move on, let's take a moment to interpret what this new tableau tells us. The basic variables (the variables corresponding to columns with a single 1 and the rest 0s) are 's', 'y', and 'R'. This means that in our current solution:

• s = 4
• y = 10
• R = 1200

The non-basic variables (the remaining variables) are 'x' and 't', which are currently equal to 0. So, our current solution is (x, y) = (0, 10), with an objective function value of R = 1200. We still have a negative entry in the 'R' row (-30 in the 'x' column), which means we can further improve our solution. This tells us that we need to perform another pivot. The process might seem a bit repetitive, but that's the beauty of the simplex method – it's a systematic way to climb towards the optimal solution! Stay tuned for the next pivot, where we'll tackle the 'x' column!

Iterating to the Optimal Solution: Second Pivot

Alright, guys, we're not done yet! Our tableau from the previous pivot looks like this:

	x	y	s	t	R	RHS
s	2	0	1	-1	0	4
t	1/2	1	0	1/2	0	10
R	-30	0	0	60	1	1200

Remember, we stopped because we still had a negative entry in the 'R' row (-30 in the 'x' column). This means we can increase R further by pivoting on the 'x' column. Let's continue our journey to the optimal solution!

Identifying the New Pivot Column and Pivot Row

• Pivot Column: The most negative entry in the 'R' row is -30, so the 'x' column is our pivot column.

• Pivot Row: We need to perform the ratio test again. Divide the RHS values by the corresponding entries in the 'x' column (only for positive entries):

  • For the 's' row: 4 / 2 = 2
  • For the 't' row: 10 / (1/2) = 20

  The smallest ratio is 2, which corresponds to the 's' row. So, the 's' row is our pivot row.

• Pivot Element: The entry at the intersection of the 'x' column and 's' row is 2. This is our new pivot element.

Pivoting: Transforming the Tableau Again

Now we pivot, just like before:

Step 1: Make the Pivot Element Equal to 1

Divide the entire pivot row (the 's' row) by the pivot element (2):

New 's' row = (Old 's' row) / 2

	x	y	s	t	R	RHS
s	1	0	1/2	-1/2	0	2
t	1/2	1	0	1/2	0	10
R	-30	0	0	60	1	1200

Step 2: Make All Other Entries in the Pivot Column Equal to 0

• Eliminate 'x' from the 't' row:

  • Subtract (1/2) times the new 's' row from the 't' row:

  New 't' row = (Old 't' row) - (1/2) * (New 's' row)

  	x	y	s	t	R	RHS
  s	1	0	1/2	-1/2	0	2
  t	0	1	-1/4	3/4	0	9
  R	-30	0	0	60	1	1200

• Eliminate 'x' from the 'R' row:

  • Add 30 times the new 's' row to the 'R' row:

  New 'R' row = (Old 'R' row) + 30 * (New 's' row)

  	x	y	s	t	R	RHS
  s	1	0	1/2	-1/2	0	2
  t	0	1	-1/4	3/4	0	9
  R	0	0	15	45	1	1260

Reached the Optimal Solution!

Look at that! Our 'R' row now has no negative entries. This is fantastic news – it means we've reached the optimal solution! We can now read off the solution from our tableau.

Interpreting the Final Tableau

Our basic variables are 'x', 'y', and 'R'. This means:

• x = 2
• y = 9
• R = 1260

The non-basic variables are 's' and 't', which are equal to 0. This tells us that our optimal solution is (x, y) = (2, 9), which gives us a maximum value of R = 1260. To put it simply, to maximize our objective function R, we should set x to 2 and y to 9. This will give us a maximum R value of 1260.

Conclusion: We Did It!

Guys, we successfully used the simplex method to solve this linear programming problem! We started with inequalities, transformed them into equations, set up our initial tableau, and then iteratively pivoted until we reached the optimal solution. It might seem like a lot of steps, but the simplex method provides a powerful and systematic way to tackle these types of optimization problems. Remember, the key is to understand the underlying principles: identifying the pivot column and row, performing the row operations, and interpreting the tableau. Once you've got those down, you can conquer any linear programming problem that comes your way! Isn't mathematics awesome? Keep practicing, and you'll become simplex method masters in no time!