Dr D’s Math Help · YouTube · 7:43
This 150-second whiteboard lesson works through the opening stage of a simplex-method solution for a two-variable maximization linear program. The instructor starts from Maximize subject to <= 78, <= 36, and x,y >= 0. She first converts the two inequality constraints into equations by adding slack variables u and w, then rewrites the objective as -5x - so the constant is zero and P stays positive. Next she builds the initial tableau with columns X, Y, U, W, P and constants on the far right. The first pivot is chosen by taking the most negative entry in the bottom row, -5 in the X column, then applying the ratio test and . Since 9 is smaller, row 2 is selected and the pivot element is 4. The clip ends as she writes the row operation () to normalize that pivot to 1. This video segment demonstrates a step in the simplex method for solving a linear programming problem. The instructor identifies the pivot element, performs row operations to update the tableau, and explains the process of converting inequality constraints into equalities using slack variables. This video segment demonstrates the final steps of solving a linear programming problem using the simplex method. The instructor performs a pivot operation on the simplex tableau, carefully explaining the row operations required to transform the matrix. After completing the pivot, she identifies that the bottom row contains no negative numbers, indicating that the optimal solution has been reached. Finally, she systematically reads the values of the decision variables (x and y) and the maximum value of the objective function (P) directly from the columns of the final tableau. This 13-second whiteboard excerpt concludes a simplex-method example for a two-variable linear programming problem. The board shows the maximization problem Maximize subject to <= 78 and <= 36, with x, y >= 0, along with several tableaus and row operations. In the clip, the instructor points to the boxed final answer and states that and , then summarizes that the solution consists of the two variables and the objective-function value. The displayed optimum is , , , , . A direct substitution check confirms that , and both constraints hold with equality, so the zero slack values indicate binding constraints at the optimum.
Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.
Generated from the video's visuals and explanation; not verbatim speech.
The clip opens on a whiteboard showing a complete maximization problem: , with constraints <= 78, <= 36, and x,y >= 0. The instructor identifies this as a linear programming problem to be solved by the simplex method, so the immediate goal is not to graph the feasible region but to convert the problem into tableau form.
She begins the conversion by handling the inequalities one at a time. For the first constraint, she adds a slack variable u, changing <= 78 into . For the second, she adds w, changing <= 36 into . The mathematical purpose is to turn each <= constraint into an equation that can occupy a row in the simplex tableau.
When she reaches the line x,y >= 0, she explicitly leaves it out of the row-by-row rewriting. Her explanation is that these nonnegativity conditions are already built into the simplex method, so they do not become separate tableau rows. This distinguishes structural assumptions of the method from the equations that must be encoded numerically.
Next she rewrites the objective. Starting from , she moves the x and y terms to the left to obtain -5x - . She emphasizes two formatting goals: there should be a zero on one side, and the coefficient of P should remain positive. That rearranged equation will become the bottom row of the tableau.
With the three equations now in hand, she constructs the initial simplex tableau. The column headers are X, Y, U, W, P, followed by a vertical separator and the constant column. The first row is filled from as 3, 5, 1, 0, 0 | 78. The second row comes from as 4, 1, 0, 1, 0 | 36. The bottom row comes from -5x - as -5, -4, 0, 0, 1 | 0. Each row is therefore just the coefficient list of one equation.
The next task is to choose the first pivot. She scans the bottom row only up to the vertical bar and looks for the largest negative number. The entry -5 under X is selected, so the X column becomes the pivot column. This step uses the objective row because negative coefficients there indicate variables that can still improve the maximization objective.
To determine the pivot row, she applies the ratio test within the chosen column. She pairs the positive entries in the X column with the constants on the far right: 78 with 3 in row 1, and 36 with 4 in row 2. The displayed ratios are 26 and 9, corresponding to and . The smallest ratio, 9, selects row 2 as the pivot row.
Because the pivot column is X and the pivot row is row 2, the pivot element is the intersection entry 4. She marks that 4 directly on the board. At this point the first pivot location has been fully determined from the initial tableau.
The final action in the clip is the start of the pivot operation itself. She writes the row operation (), explaining that the whole second row will be divided by 4 so the pivot entry becomes 1. This normalization is the preparatory step before eliminating the other entries in the pivot column; the fully updated row is not completed within this excerpt.
The instructor begins by explaining the current state of the simplex tableau, noting the values obtained after previous operations.
The focus shifts to identifying the pivot element, which is crucial for the next set of row operations aimed at creating zeros above and below it in the tableau.
Detailed instructions are given on how to perform the row operations: subtracting three times the second row from the first row and adding five times the second row to the third row.
As the calculations proceed, the instructor updates the tableau on the whiteboard, showing the new values resulting from the row operations.
The process is reiterated, emphasizing the importance of selecting the correct pivot element based on the most negative indicator in the bottom row of the tableau.
Further calculations involve dividing specific values to determine the next pivot element, ensuring the tableau moves closer to the optimal solution.
The instructor begins by explaining the row operation needed to normalize the pivot row. She states that multiplying the first row by is equivalent to dividing by the pivot element, which is . This step ensures the pivot element becomes 1.
She then writes out the new values for the first row after the division: 0, 1, , -3/17, 0, and 12. She notes that the other rows will remain unchanged during this specific step.
Moving to the next phase of the pivot, the instructor explains that the goal is to create zeros in the rest of the pivot column. She identifies the pivot element again and points to the entries in the column that need to be eliminated.
To eliminate the in the second row, she specifies the row operation . To eliminate the -11/4 in the third row, she specifies the row operation .
The instructor confirms that the first row remains unchanged as it is the pivot row. She then calculates and writes the new second row: 1, 0, -1/17, , 0, and 6.
Finally, she calculates and writes the new third row: 0, 0, , , 1, and 78. The tableau is now fully updated for this iteration.
With the new tableau complete, the instructor checks the bottom row for any negative numbers. Finding none, she declares that the optimal solution has been reached and the simplex method is finished.
She explains how to read the solution from the final tableau. Columns containing a single 1 and 0s elsewhere correspond to the basic variables. The value of the variable is the number in the rightmost column of the row containing the 1.
Identifying the column for x, she sees the 1 is in the second row, so . For y, the 1 is in the first row, so . For the objective function P, the 1 is in the third row, so .
She concludes by stating that any columns without a single 1 and 0s elsewhere represent non-basic variables, which are set to zero. Thus, the slack variables u and w are both zero.
The clip opens on a whiteboard already filled with a linear-programming example. At the upper left, the problem is written as Maximize subject to , , and . To the right and below are simplex tableaus and row-operation notes, and at the far right a boxed final answer lists , , , , and . As the instructor points to the box, she says, "u equals zero, w equals zero here." Mathematically, and are the slack variables introduced when the inequalities were rewritten as equations, so their being zero means both original constraints are met exactly at equality in the optimal solution.
The instructor then summarizes the result verbally: "Okay, and we have our solution, the two variables and the value of the objective function for this linear programming problem." The key takeaway is that the completed simplex tableau has yielded the optimal decision-variable values and , together with the objective value . This can be checked directly from the stated objective function: substituting the solution gives , which matches the boxed value on the board. The same substitution also shows why the slack variables vanish: and , so both constraints are binding at the optimum.
The worked example begins with a two-variable maximization problem. The objective is , and the constraints are <= 78 and <= 36, together with x,y >= 0. The instructor frames this as the input to the simplex method.
Each <= constraint is rewritten by adding a nonnegative slack variable. The first constraint becomes , and the second becomes . This is the key preprocessing step that makes tableau construction possible.
The conditions x,y >= 0 remain part of the problem assumptions but are not turned into extra equations in the displayed setup. The instructor states that they are already included in the simplex method framework.
Before forming the tableau, the objective is rearranged to -5x - . The instructor specifically keeps P positive and places 0 on the other side so the objective can serve as the bottom row.
The tableau columns are X, Y, U, W, P, with constants separated on the far right. The first two rows come from the constraint equations and the last row comes from the rewritten objective equation.
To choose the pivot column, inspect the bottom row before the constant separator and select the most negative entry. Here -5 under X is the largest negative number, so the X column is the pivot column.
After fixing the pivot column, divide the constants by the corresponding positive entries in that column. The displayed ratios are and . The smaller ratio, 9, chooses row 2 as the pivot row.
The pivot element is where the chosen pivot column and pivot row intersect. Since the pivot column is X and the pivot row is row 2, the pivot element is the entry 4.
The first row operation in the pivot step is to scale the entire pivot row so the pivot element becomes 1. The instructor writes (), meaning divide row 2 by 4.
The simplex method is an algorithm used to solve linear programming problems by iteratively improving the candidate solution until an optimal solution is found.
The pivot element is chosen based on the most negative indicator in the bottom row of the simplex tableau, guiding the row operations to update the tableau.
Row operations such as adding multiples of rows are used to create zeros above and below the pivot element, transforming the tableau towards the optimal solution.
Slack variables are added to inequality constraints to convert them into equality constraints, enabling the application of the simplex method.
The simplex method is an algorithm for solving linear programming problems. It works by moving from one basic feasible solution to another, improving the objective function value at each step, until an optimal solution is found.
Pivoting is the core operation in the simplex method. It involves selecting a pivot element, normalizing its row, and using row operations to create zeros in the rest of the pivot column. This moves the solution to an adjacent vertex of the feasible region.
For a maximization problem, the simplex method terminates when there are no negative numbers in the bottom row (the objective function row) of the tableau. This indicates that no further improvement in the objective function is possible.
In the final tableau, basic variables correspond to columns with a single 1 and 0s elsewhere. Their values are found in the rightmost column. Non-basic variables (columns without this pattern) are set to zero. The objective function value is read similarly.
The completed whiteboard work reports the optimal solution as , , and . The instructor explicitly calls this "our solution, the two variables and the value of the objective function for this linear programming problem." This card captures the end result of the simplex procedure shown on the board.
The variables and are slack variables added to turn the constraints into equations. In the final answer they are both zero, which means neither constraint has leftover slack at the optimum. Equivalently, both constraints hold with equality at the optimal point.
Using the stated objective function , substitute the reported optimal decision variables. This gives , confirming the boxed objective value shown on the board.
Substitution also verifies the constraint activity: and . Since both left-hand sides equal their right-hand sides, the optimal solution lies at the intersection of the two constraint boundaries, consistent with zero slack variables.
Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.
Whiteboard shows the objective function and later the rewritten form -5x - .
The instructor says to rewrite the objective function with a zero on one side and wants the P to be positive.
P
Objective value to be maximized.
Real-valued objective variable; appears as the last column in the tableau.
Whiteboard shows x in the constraints <= 78, <= 36, and x,y >= 0.
The initial tableau header lists X as the first decision-variable column.
x
First decision variable in the linear programming problem.
Nonnegative real variable, constrained by x >= 0.
Whiteboard shows y in the constraints <= 78, <= 36, and x,y >= 0.
The initial tableau header lists Y as the second decision-variable column.
y
Second decision variable in the linear programming problem.
Nonnegative real variable, constrained by y >= 0.
The instructor says, 'For our first constraint, we're going to insert the slack variable u to turn it into an equation instead of an inequality.'
Whiteboard shows and the tableau column U.
u
Slack variable added to the first constraint to convert the inequality into an equation.
Nonnegative slack variable associated with <= 78.
The instructor says, 'For our second, we'll insert the slack variable w to turn it into an equation instead of an inequality.'
Whiteboard shows and the tableau column W.
w
Slack variable added to the second constraint to convert the inequality into an equation.
Nonnegative slack variable associated with <= 36.
The instructor writes the tableau header X, Y, U, W, P and then a vertical separator before the constant column.
She says the variables are listed across the top and each row is the coefficients of one of those equations.
X, Y, U, W, P
Column labels of the initial simplex tableau, followed by the rightmost constant column.
Tableau headers for the two constraints and the objective row.
Whiteboard shows () beside the pivot row.
The instructor says to replace row two with one-fourth row two, dividing the whole row by four.
()
Row operation multiplying the second constraint row by so the pivot element becomes 1.
Applied to the second row of the initial tableau.
The objective function is written as .
P
Objective function value to be maximized.
Real numbers
The variable x appears in the objective function and constraints.
x
Decision variable.
Non-negative real numbers
The variable y appears in the objective function and constraints.
y
Decision variable.
Non-negative real numbers
The slack variable u is added to the first constraint: .
u
Slack variable for the first constraint.
Non-negative real numbers
The slack variable w is added to the second constraint: .
w
Slack variable for the second constraint.
Non-negative real numbers
The instructor states they will solve a linear programming problem using the simplex method.
Whiteboard shows Maximize subject to <= 78, <= 36, x,y >= 0.
The clip begins with a standard maximization linear program: maximize an objective function in two nonnegative variables subject to two linear inequality constraints. The instructor identifies this as the problem to be solved by the simplex method.
The problem is a maximization linear program.
There are two decision variables, x and y.
There are two inequality constraints plus nonnegativity restrictions.
The instructor says to insert the slack variable u to turn the first constraint into an equation instead of an inequality.
Whiteboard shows .
To prepare the first inequality for the simplex tableau, the instructor adds a slack variable u. This changes the inequality <= 78 into the equality , preserving the same feasible region when u is nonnegative.
The original constraint is <= 78.
u is introduced as a slack variable.
The inequality direction is <=, so adding a nonnegative slack converts it to equality.
The instructor says to insert the slack variable w to turn the second constraint into an equation instead of an inequality.
Whiteboard shows .
The second inequality is converted similarly by adding a slack variable w, producing the equality . This gives the second row needed for the initial simplex tableau.
The original constraint is <= 36.
w is introduced as a slack variable.
The inequality direction is <=, so adding a nonnegative slack converts it to equality.
The instructor says, 'We don't need to worry about these ones because they're included in the simplex method.'
The whiteboard still shows x,y >= 0 in the original problem statement.
The instructor explicitly does not create separate tableau rows for x,y >= 0. Instead, she states that these conditions are already included in the simplex method framework, so only the two main constraints and the objective are rewritten for the tableau.
The original problem includes nonnegativity restrictions on x and y.
These restrictions are not rewritten as separate equations in the displayed tableau setup.
The instructor says to move everything over to the left-hand side because they want a zero on one side and want the P to be positive.
Whiteboard shows -5x - .
The objective is rearranged so all terms are on one side and the constant is zero. The instructor specifically keeps P positive, yielding -5x - , which becomes the bottom row of the initial tableau.
Start from .
Move the x and y terms to the left side.
Keep the coefficient of P equal to +1.
The instructor says, 'So now we have our initial tableau,' then explains that the variables are listed across the top and each row is the coefficients of one of those equations.
The tableau is written with headers X, Y, U, W, P and rows 3 5 1 0 0 | 78, 4 1 0 1 0 | 36, and -5 -4 0 0 1 | 0.
After rewriting the constraints and objective as equations, the coefficients are arranged into a tableau. The columns correspond to x, y, u, w, and P, while the rightmost column holds constants. The first two rows come from the constraints and the last row comes from the rewritten objective.
Use the equations , , and -5x - .
Place variable coefficients under matching column headers.
Put constants in the far-right column.
The instructor says to go across the bottom row until the vertical bar and choose the largest negative number, which chooses the pivot column.
She circles the -5 in the bottom row under the X column.
For this maximization tableau, the pivot column is selected by scanning the bottom row up to the vertical separator and choosing the most negative entry. In the displayed tableau, -5 is the largest negative number, so the X column becomes the pivot column.
Apply to the bottom objective row of the initial tableau.
Only entries before the vertical constant separator are considered.
The chosen entry here is -5 in the X column.
The instructor says to take the numbers in this column and divide them into the numbers in the constant column, meaning the numbers on the far right side.
Whiteboard shows 26 and 9 written to the right of the first two rows.
The ratios are computed from the pivot column entries 3 and 4 against the constants 78 and 36.
The spoken phrase 'divide them into' is ambiguous in ordinary English, but the displayed arithmetic matches constant divided by pivot-column entry.
Once the pivot column is fixed, the instructor forms ratios using the positive entries in that column and the corresponding constants on the far right. For this tableau the ratios are and . The smallest ratio determines the pivot row.
Use the entries in the pivot column from the constraint rows.
Pair each with the constant in the same row.
Select the row with the smallest nonnegative ratio.
The instructor asks which of those numbers is smallest, answers nine, and says that chooses the pivot row; then she says the four is our pivot element.
The 4 in the second row and first column is circled.
The pivot element is the intersection of the chosen pivot column and pivot row. Here the X column was selected from the bottom row, and the second constraint row was selected by the ratio test, so the entry 4 at their intersection is the pivot element.
Pivot column is the X column.
Pivot row is the second constraint row because its ratio 9 is smallest.
The pivot element is the coefficient where they meet.
The instructor says the first thing to do is use row operations to make that a one, replacing row two with one-fourth row two.
Whiteboard shows ().
She points to the circled 4 and indicates the whole second row is divided by 4.
The updated row values after division are not fully written out within this clip.
The next simplex step is to turn the pivot element into 1 by scaling its entire row. The instructor applies the operation (), meaning every entry in row 2 is divided by 4. This prepares the tableau for eliminating the other entries in the pivot column.
The pivot element is 4 in row 2, column X.
Multiply the entire second row by .
This makes the pivot entry equal to 1.
The instructor explains the steps of the simplex method, including identifying pivot elements and performing row operations.
The simplex method is an algorithm for solving linear programming problems. It involves iteratively moving along the edges of the feasible region to find the optimal solution by selecting pivot elements and performing row operations.
Linear objective function
Linear inequality constraints
Non-negativity constraints
The instructor identifies the pivot element and explains its role in row operations.
In the simplex method, the pivot element is the entry in the tableau that is used to perform row operations to create zeros above and below it in its column.
Must be non-zero
Typically chosen based on specific rules such as the most negative indicator in the bottom row
The instructor says, 'Now, we have completed this because we have no more negative numbers on the bottom row.'
For a maximization linear programming problem, if the bottom row of the simplex tableau contains no negative numbers, the current basic feasible solution is optimal.
The problem is a maximization linear programming problem.
The tableau represents a basic feasible solution.
For all maximization linear programming problems.
The instructor sequentially rewrites the constraints with slack variables, rewrites the objective, and then reads off the tableau rows.
Whiteboard shows , , -5x - , followed by the tableau rows.
Start from the first inequality constraint of the original problem.
Given on the whiteboard as part of the linear programming statement.
Add the slack variable u to convert the first inequality into an equation.
Stated by the instructor as the first step of the simplex setup.
Start from the second inequality constraint.
Given on the whiteboard as part of the original problem.
Add the slack variable w to convert the second inequality into an equation.
Stated by the instructor as the analogous second setup step.
Begin with the objective function to be maximized.
Given on the whiteboard.
Move all terms to the left side while keeping P positive.
The instructor explicitly says they want a zero on one side and P positive.
Place the coefficients of the three equations into a tableau with constants on the far right.
The instructor says each row is the coefficients of one of those equations and writes the initial tableau.
The original maximization problem is transformed into the initial simplex tableau used for pivoting.
The instructor explains the pivot-column rule, computes the ratios, compares 26 and 9, and identifies 4 as the pivot element.
The -5 is circled, the ratios 26 and 9 are written, and the 4 is circled.
Inspect the objective row before the constant column.
The instructor says to go across the bottom row until the vertical bar.
Choose the largest negative number in that row.
Stated directly by the instructor as the pivot-column rule.
Because -5 lies under the X column, the X column is selected.
Follows from the location of the circled -5 in the tableau.
Form the ratio for the first constraint row using its pivot-column entry 3 and constant 78.
The instructor says to divide the column numbers into the constant-column numbers and writes 26.
Form the ratio for the second constraint row using its pivot-column entry 4 and constant 36.
The instructor writes 9 beside the second row during the ratio test.
Compare the two ratios and select the smaller one.
The instructor asks which number is smallest and answers nine.
The second row is chosen, so the intersection with the X column is the entry 4.
Explicitly stated by the instructor after the ratio comparison.
The first pivot is the entry 4 in row 2, column X.
The instructor says to use row operations to make the pivot a one by replacing row two with one-fourth row two.
Whiteboard shows ().
The resulting normalized row is announced conceptually but not fully completed onscreen in this clip.
Identify the current pivot from the previous selection step.
Established immediately before this derivation in the video.
Multiply the entire second row by .
The instructor explicitly names this row operation.
The pivot entry becomes 1 after scaling.
Arithmetic consequence of the stated row operation.
Applying () normalizes the pivot element to 1, preparing the next elimination steps.
The instructor explains and performs row operations to update the tableau.
To make the element in the first row and second column zero, subtract three times the second row from the first row.
Row operation rule
To make the element in the third row and second column zero, add five times the second row to the third row.
Row operation rule
The updated tableau has zeros above and below the pivot element in the second column.
The instructor explicitly states the row operations: 'replace the first row by times itself', '', and ''.
Multiply the first row by to make the pivot element 1.
Elementary row operation to normalize the pivot row.
Subtract of the new first row from the second row to make the element in the pivot column zero.
Elementary row operation to eliminate the entry above/below the pivot.
Add of the new first row to the third row to make the element in the pivot column zero.
Elementary row operation to eliminate the entry above/below the pivot.
The tableau is transformed into the next iteration, with the pivot column now having a 1 in the pivot position and 0s elsewhere.
The board states Maximize .
The boxed final answer lists , , and .
Start from the objective function written on the board.
Observed directly from the whiteboard statement of the maximization problem.
Substitute the reported optimal decision-variable values into the objective function.
The final boxed solution on the board gives and .
Evaluate each product separately.
Arithmetic substitution step.
Add the two terms to obtain the objective value.
Arithmetic simplification; this matches the boxed value on the board.
The displayed objective value is consistent with the stated objective function when and are substituted.
The instructor frames the clip as solving a linear programming problem using the simplex method.
The full worked setup appears on the whiteboard from the original inequalities through the first pivot operation.
The example is only carried through the first normalization step; later simplex iterations are outside this clip.
Maximize subject to <= 78, <= 36, x,y >= 0, using the simplex method.
Objective: .
Constraint 1: <= 78.
Constraint 2: <= 36.
Nonnegativity: x,y >= 0.
Rewrite the problem for the simplex method, construct the initial tableau, choose the first pivot, and begin the pivot row normalization.
Introduce slack variable u in the first constraint.
Converts the first inequality into equation form for the tableau.
Introduce slack variable w in the second constraint.
Converts the second inequality into equation form for the tableau.
Rewrite the objective with all terms on one side and P positive.
Required to place the objective as the bottom row of the tableau.
Assemble the initial simplex tableau from the three equations.
Each row contains the coefficients of one equation, with constants on the far right.
Select the pivot column by taking the most negative entry in the bottom row before the constant column.
This is the rule stated by the instructor for the first pivot choice.
Compute the ratios for the constraint rows in the pivot column.
The ratio test uses the pivot-column entries and the constants on the far right.
Select the pivot row from the smallest ratio.
Stated directly by the instructor after comparing the two ratios.
Normalize the pivot row so the pivot element 4 becomes 1.
This is the first row operation announced for the pivot step.
The example reaches the initial tableau, identifies the pivot element 4 in row 2 column X, and begins the simplex iteration with the row operation ().
The displayed tableau rows match the rewritten equations, the circled -5 and 4 match the stated pivot rules, and the written ratios 26 and 9 agree with and .
The instructor walks through a step of the simplex method using a specific example.
Maximize subject to , , x, .
Objective function:
Constraints: , , x,
Find the values of x and y that maximize P.
Convert the first inequality constraint into an equality by adding the slack variable u.
Definition of slack variables
Convert the second inequality constraint into an equality by adding the slack variable w.
Definition of slack variables
Rewrite the objective function in standard form for the simplex tableau.
Standard form for simplex method
Select the pivot element based on the most negative indicator in the bottom row.
Simplex method rule
Use row operations to create zeros above and below the pivot element.
Row operation rule
The updated tableau after performing row operations.
Check that the pivot column has zeros above and below the pivot element.
The problem is stated as: Maximize subject to <= 78, <= 36, x, y >= 0.
Maximize subject to <= 78, <= 36, x, y >= 0.
Objective function:
Constraint 1: <= 78
Constraint 2: <= 36
Non-negativity constraints: x, y >= 0
Find the values of x and y that maximize P while satisfying all constraints.
Set up the initial simplex tableau with slack variables u and w.
Standard procedure for converting inequalities to equations for the simplex method.
Perform the first pivot operation on the element 4 in the second row, first column.
Selecting the most negative indicator in the bottom row (-5) and the smallest non-negative ratio ().
Perform the second pivot operation on the element 1 in the first row, second column.
Selecting the most negative indicator in the bottom row (-11/4) and the smallest non-negative ratio ().
Read the optimal solution from the final tableau.
Columns with a single 1 and 0s elsewhere correspond to basic variables; their values are in the rightmost column. Non-basic variables are set to 0.
The maximum value of P is 78, which occurs when and .
Substitute and into the original constraints and objective function to verify they are satisfied and yield .
The board shows the problem: Maximize subject to <= 78, <= 36, x, y >= 0.
The board shows multiple simplex tableaus and row-operation annotations leading to a boxed final answer.
The speaker concludes, "And we have our solution, the two variables and the value of the objective function for this linear programming problem."
The clip begins after earlier setup steps, so not every intermediate tableau transformation is fully narrated within this excerpt.
Solve the linear programming problem shown on the board and identify the optimal values of the decision variables, objective function, and slack variables.
Maximize .
Constraints: <= 78 and <= 36.
Nonnegativity: x, y >= 0.
Slack variables u and w are introduced in the tableau.
The final boxed answer on the board is , , , , .
Read off the optimal solution from the completed simplex tableau and state the values of x, y, P, u, and w.
The first inequality is represented in equation form using slack variable u.
Visible in the converted-system writing on the board.
The second inequality is represented in equation form using slack variable w.
Visible in the converted-system writing on the board.
The objective function is rewritten so that P can be tracked in the tableau.
Visible in the converted-system writing on the board.
From the completed tableau, the instructor records the final basic-variable values and the objective value in a boxed answer.
Directly observed in the boxed solution on the right side of the board and reinforced by the speaker's concluding statement.
, , , , .
Substituting and into gives , matching the reported objective value. Substituting into the constraints gives and , so both constraints hold with equality, consistent with and .
A whiteboard shows the original maximization problem in stacked lines before the instructor begins writing the transformed equations.
Whiteboard text: Maximize
Constraint <= 78
Constraint <= 36
Nonnegativity x,y >= 0
No new writing yet; the original problem remains fixed while the instructor introduces the simplex method.
The objective and both constraints stay visible unchanged during the introduction.
This visual establishes the starting linear programming problem before any simplex preprocessing.
The instructor writes the transformed equations one by one to the right of the original problem.
The new lines , , and -5x - appear sequentially.
Original inequalities on the left
New equation
New equation
New equation -5x -
First the left constraint is rewritten with u.
Then the right constraint is rewritten with w.
Finally the objective is moved to the form with zero on one side.
The original problem statement remains visible while the transformed system is added beside it.
The visual sequence shows the algebraic preprocessing required before constructing the simplex tableau.
The instructor draws a table, writes the headers X, Y, U, W, P, then fills the rows with coefficients and constants.
The completed initial tableau is visible with a vertical separator before the constant column.
Table grid
Column headers X, Y, U, W, P
Row 1 coefficients 3, 5, 1, 0, 0 and constant 78
Row 2 coefficients 4, 1, 0, 1, 0 and constant 36
Objective row coefficients -5, -4, 0, 0, 1 and constant 0
The header row is written first.
Then the first constraint row is filled.
Then the second constraint row is filled.
Then the objective row is filled along the bottom.
The column order remains X, Y, U, W, P throughout the construction.
The constants remain separated by the vertical bar on the far right.
This animation converts the rewritten equations into the standard tableau layout used for pivoting.
The instructor circles the -5 in the bottom row, writes the ratios 26 and 9 to the right of the first two rows, and then circles the 4 in row 2.
Circled -5 in the bottom row
Written ratio 26 beside row 1
Written ratio 9 beside row 2
Circled 4 in row 2, column X
The bottom-row candidate -5 is highlighted first.
The ratio values are added next to the constraint rows.
The winning pivot element 4 is circled last.
The underlying tableau entries do not change during the selection phase.
The X column remains the chosen pivot column once -5 is circled.
The markings show how the pivot column, pivot row, and pivot element are identified from the tableau.
The notation () is written beside the pivot row.
The instructor points between the circled 4 and the new row-operation note.
The fully updated second row after division is not shown within this clip.
Circled pivot element 4
Notation ()
Second tableau row
A row-operation label is added to indicate scaling the entire second row by .
The pivot remains in row 2, column X.
The rest of the tableau is still in its initial form during this annotation.
This visual records the first elementary row operation used to normalize the pivot to 1.
The tableau is updated on the whiteboard as row operations are performed.
Simplex tableau
Pivot element
Row operations
Values in the tableau change as row operations are applied.
Zeros are created above and below the pivot element.
Structure of the tableau remains the same.
Variables and constraints remain unchanged.
The visual updates demonstrate the iterative nature of the simplex method and how row operations transform the tableau towards the optimal solution.
The instructor writes the new rows of the tableau on the whiteboard after performing the row operations.
Initial simplex tableau
Row operations
New simplex tableau
The values in the rows of the tableau change according to the specified row operations.
The pivot column is transformed to have a 1 in the pivot position and 0s elsewhere.
The structure of the tableau (rows and columns representing variables and constants) remains the same.
The system of linear equations represented by the tableau remains equivalent.
The visual transformation demonstrates the mechanical process of the simplex method, where algebraic row operations correspond to moving from one basic feasible solution to another, improving the objective function value.
The instructor stands at the right side of the whiteboard and points toward the boxed final answer while speaking.
The boxed answer contains , , , , .
Boxed final answer on the right side of the board
Instructor's hand and marker
Multiple simplex tableaus across the board
The instructor gestures toward the boxed solution while verbally identifying it as the answer.
Attention shifts from the tableau work to the extracted variable values.
The written values in the box remain unchanged during the clip.
The objective function and constraints already written on the board remain visible.
The visual emphasis marks the transition from tableau computation to solution interpretation: the optimal decision variables, objective value, and slack values are being read off the completed simplex work.
The instructor says they do not need to worry about these ones because they are included in the simplex method.
One might think every displayed condition, including x,y >= 0, must be rewritten as its own equation or tableau row.
In this setup, the nonnegativity conditions are treated as built-in assumptions of the simplex method rather than separate rows in the initial tableau.
The instructor says to take the numbers in this column and divide them into the numbers in the constant column.
The written results 26 and 9 correspond to and .
The spoken wording is potentially confusing even though the displayed arithmetic is clear.
The phrase 'divide them into' can be misread as dividing the constant by the wrong quantity or reversing the ratio.
The displayed computations show the intended ratio is constant divided by the positive pivot-column entry, giving and .
The speaker explicitly states, "u equals zero, w equals zero here."
The final boxed answer includes and .
Learners may think that because and , the corresponding constraints are irrelevant or unused.
In fact, zero slack means the opposite: those constraints are active at equality at the optimum. The constraints still shape the feasible region; they are simply binding at the chosen solution point.
The instructor first names the linear programming problem and then immediately says the first step is to insert slack variables.
The inequalities are rewritten as equations with u and w.
The general linear programming problem is prepared for the simplex method by applying slack-variable conversion to its inequality constraints.
The equations with u and w are directly transcribed into the first two tableau rows.
The initial tableau depends on having already rewritten the constraints as equations with slack variables.
The rewritten objective -5x - becomes the bottom row -5, -4, 0, 0, 1 | 0.
The bottom row of the initial tableau is obtained from the rearranged objective equation.
The pivot-column rule is applied directly to the bottom row of the displayed tableau.
The pivot-column selection rule is applied to the completed initial tableau.
After choosing the X column, the instructor says to take the numbers in this column and divide them into the constant-column numbers.
The ratio test cannot be performed until the pivot column has been selected.
The smallest ratio 9 chooses the pivot row, and then the instructor identifies the 4 as the pivot element.
The pivot element is determined by combining the chosen pivot column with the pivot row selected by the ratio test.
Once the 4 is identified as the pivot, the instructor writes () to normalize it.
The row operation is chosen specifically to turn the identified pivot element into 1.
The instructor uses row operations as part of the simplex method.
Row operations are a fundamental tool used within the simplex method to manipulate the tableau and find the optimal solution.
The instructor identifies the pivot element and then performs row operations around it.
The selection of the pivot element dictates which row operations are performed to update the tableau.
The instructor uses pivoting as a step within the broader simplex method.
Pivoting is a core iterative step contained within the overall procedure of the simplex method.
The instructor concludes the simplex method by reading the final tableau.
Reading the final tableau is the concluding step of the simplex method to extract the optimal solution.
The board shows constraints <= 78 and <= 36 together with slack variables u and w.
The speaker says, "u equals zero, w equals zero here."
The interpretation of zero slack variables applies directly to the example's final solution, showing that both original constraints are tight at the optimum.
The instructor says the first step is to rewrite the problem inserting slack variables and rewrite the objective function with a zero on one side.
The instructor says the slack variables turn the inequalities into equations instead of inequalities.
The instructor says they do not need to worry about these ones because they are included in the simplex method.
The objective is rewritten as -5x - .
The completed initial tableau is shown with headers and three rows of coefficients.
The instructor says to choose the largest negative number in the bottom row up to the vertical bar.
The written ratios are 26 and 9, corresponding to and .
The instructor says the smallest ratio chooses the pivot row and the four is the pivot element.
The board shows () after the pivot is identified.
The instructor explains how to select the pivot element.
The instructor performs and explains row operations.
The instructor explains the row operations required for pivoting.
Covered · Introduction and display of the original maximization linear program.
Covered · Both inequality constraints are rewritten with slack variables u and w.
Covered · The instructor explains that x,y >= 0 are not rewritten as separate tableau rows.
Covered · The objective function is rearranged to -5x - .
Covered · The initial simplex tableau is drawn and filled from the rewritten equations.
Covered · The pivot column is chosen from the most negative bottom-row entry, -5 in the X column.
Covered · The ratio test is performed, producing 26 and 9.
Covered · The smallest ratio selects row 2, and the pivot element is identified as 4.
Covered · The instructor announces the row operation () to normalize the pivot to 1.
Covered · All mathematical content within the segment is covered.
Covered · The instructor performs the second pivot operation, explaining the row operations and writing the resulting tableau.
Covered · The instructor identifies the final tableau, explains the optimality condition, and reads off the values of the decision variables and the objective function.
Covered · Audio states and while the board shows the boxed final answer.
Covered · The speaker identifies the solution as the two variables plus the objective-function value; the board continues to display the full simplex setup and final boxed result.
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