Duality Problem 1,2 - Linear Programming Problems (LPP) - Engineering Mathematics - 4

1. Formula
   Observation

   Whiteboard shows objective functions written as "Max Z = ..." and "Min Z = ...".

Z

Objective function value in the linear programming problems.

Real-valued objective value; the video does not state units.

1. Formula
   Observation

   Both displayed primal problems use variables $x_1$ and $x_2$ in the objective and constraints.

$x_1$, $x_2$

Decision variables of the two displayed primal linear programs.

The board explicitly states $x_1$, $x_2 \ge 0$ for both problems.

1. Formula
   Observation

   Top-right note box contains "Max -> <=" and "Min -> >=".

2. Audio
   Observation

   The instructor says maximization problems should have all constraints less than equal to, and minimization problems should have all constraints greater than equal to.

Max, Min

Optimization direction labels used to distinguish maximization-type and minimization-type problems.

Applied here to primal problems and their corresponding duals.

1. Audio
   Observation

   The instructor repeatedly contrasts "primal" and "dual" and says the given problem is primal while its solution is dual.

primal, dual

Terminology for the given optimization problem and the associated transformed problem to be formed.

Used in the context of linear programming duality.

1. Formula
   Observation

   Whiteboard shows the primal objective as Max $Z = 40x_1 + 35x_2$.

Z

Objective value of the primal maximization problem.

Real-valued objective function.

1. Formula
   Observation

   The primal uses variables $x_1$ and $x_2$ in the objective and constraints.

2. Audio
   Observation

   The instructor refers to coefficients of $x_1$ and $x_2$ when forming dual constraints.

$x_1$

First decision variable of the primal linear program.

Nonnegative real variable, $x_1 \ge 0$.

1. Formula
   Observation

   The primal uses variables $x_1$ and $x_2$ in the objective and constraints.

2. Audio
   Observation

   The instructor says there are two variables, $x_1$ and $x_2$, hence two dual constraints.

$x_2$

Second decision variable of the primal linear program.

Nonnegative real variable, $x_2 \ge 0$.

1. Formula
   Observation

   The instructor writes Min $W = 60y_1 + 96y_2$ under 'Dual is given by'.

2. Audio
   Observation

   He states that if the primal is maximize, its dual will become minimize.

W

Objective value of the dual minimization problem.

Real-valued objective function.

1. Formula
   Observation

   The dual objective contains $60y_1$, and the dual constraints contain $2y_1$ and $3y_1$.

2. Audio
   Observation

   The instructor says the dual objective is formed from the right-hand side constants 60 and 96.

$y_1$

Dual variable associated with the first primal constraint $2x_1 + 3x_2 \le 60$.

Nonnegative real variable, $y_1 \ge 0$.

1. Formula
   Observation

   The dual objective contains $96y_2$, and the dual constraints contain $4y_2$ and $8y_2$.

2. Audio
   Observation

   The instructor names 96 as the second right-hand-side constant and pairs it with $y_2$.

$y_2$

Dual variable associated with the second primal constraint $4x_1 + 8x_2 \le 96$.

Nonnegative real variable, $y_2 \ge 0$.

1. Formula
   Observation

   $x_1$ appears in the primal objective and constraints.

$x_1$

Decision variable of the primal problem.

$x_1 \ge 0$

1. Formula
   Observation

   $x_2$ appears in the primal objective and constraints.

$x_2$

Decision variable of the primal problem.

$x_2 \ge 0$

1. Audio
   Observation

   Opening speech introduces "duality, primal and dual problem number one and two".

2. Formula
   Observation

   Board heading reads "Primal & Dual [Duality]".

This clip introduces duality in linear programming by naming the topic as "Primal & Dual [Duality]" and stating that the task is to find the dual of given problems. The instructor also distinguishes this topic from the separate method called "dual simplex method".

1. The discussion concerns forming a dual problem from a given primal linear program.

2. The instructor explicitly says this is different from the dual simplex method.

1. Audio
   Observation

   The instructor says if the problem is of maximization type, all the constraints will be less than equal to.

2. Formula
   Observation

   Top-right note box shows "Max -> <=".

For a maximization-type linear programming problem in the form used here, every constraint is required to be of the form $\le$. The instructor states this applies when checking whether the primal is in the correct form before forming the dual.

1. Applies to the maximization-type problem under discussion.

2. Stated for primal and dual alike in the instructor's wording.

1. Topic introduction: primal and dual in linear programming

1. Audio
   Observation

   The instructor says whenever you get minimization type problem, all the constraints should be greater than equal to.

2. Formula
   Observation

   Top-right note box shows "Min -> >=".

For a minimization-type linear programming problem in the form used here, every constraint is required to be of the form $\ge$. This is presented as the counterpart to the maximization rule.

1. Applies to the minimization-type problem under discussion.

2. Stated for primal and dual alike in the instructor's wording.

1. Topic introduction: primal and dual in linear programming

1. Audio
   Observation

   The instructor says if the problem is of maximization, that is your primal, its dual is minimize and vice versa.

The clip states that if the primal problem is a maximization problem, then its dual is a minimization problem, and conversely if the primal is minimization then the dual is maximization. This is presented as part of the basic setup for forming the dual.

1. Used when converting a primal linear program into its dual.

2. The clip does not yet show the full coefficient-transposition rule.

1. Topic introduction: primal and dual in linear programming

1. Audio
   Observation

   The instructor says the problem is maximize and all constraints are less than equal to, so the primal is in the correct form.

2. Formula
   Observation

   The board shows Max $Z = 40x_1 + 35x_2$ with sub to $2x_1 + 3x_2 \le 60$, $4x_1 + 8x_2 \le 96$, $x_1$,$x_2 \ge 0$.

For this lesson's rule set, a primal problem can be converted directly into its dual when it is written as a maximization problem whose functional constraints use $\le$ and whose variables are nonnegative. The instructor explicitly checks these conditions before writing the dual.

1. The primal objective is a maximization.

2. All functional constraints are of the form $\le$.

3. All decision variables are nonnegative.

1. Audio
   Observation

   The instructor says the objective function will be formed based on the right hand side of the constraint.

2. Formula
   Observation

   He writes Min $W = 60y_1 + 96y_2$ after pointing to the constants 60 and 96.

When the primal has two $\le$ constraints with right-hand sides 60 and 96, the dual minimization objective uses those numbers as coefficients of the new dual variables $y_1$ and $y_2$. Thus the dual objective is built from the primal constraint constants rather than from the primal objective coefficients.

1. Applied to the displayed primal maximization problem with two constraints.

2. Each primal constraint contributes one dual variable.

1. Correct form for constructing the dual

1. Audio
   Observation

   The instructor says the constraints will be formed based on the coefficient of $x_1$, then later based on the coefficient of $x_2$.

2. Formula
   Observation

   He writes $2y_1 + 4y_2 \ge 40$ and $3y_1 + 8y_2 \ge 35$.

Each primal decision variable produces one dual constraint. For $x_1$, the coefficients down the primal constraints are 2 and 4, giving $2y_1 + 4y_2$; the right-hand side is the primal objective coefficient 40. For $x_2$, the coefficients are 3 and 8, giving $3y_1 + 8y_2$; the right-hand side is the primal objective coefficient 35.

1. Used here for a primal maximization problem in the stated correct form.

2. One dual constraint is created per primal variable.

1. Correct form for constructing the dual
2. Dual objective coefficients come from primal right-hand sides

1. Audio
   Observation

   The instructor says since it is minimize, for minimize all the constraints should be greater than equal to.

2. Formula
   Observation

   The dual constraints are written with $\ge$ signs: $2y_1 + 4y_2 \ge 40$ and $3y_1 + 8y_2 \ge 35$.

In this worked example, once the primal maximization problem is converted to a dual minimization problem, the functional constraints of the dual are written using $\ge$. The instructor treats this as part of the standard recipe for forming the dual from the given primal form.

1. Stated for the dual being constructed here from a primal max problem with $\le$ constraints.

1. Correct form for constructing the dual

1. Audio
   Observation

   The instructor says finally $y_1$, $y_2$ greater than equal to 0. This is your by default.

2. Formula
   Observation

   He writes $y_1$, $y_2 \ge 0$ at the bottom of the dual.

After constructing the dual objective and constraints, the instructor appends $y_1$, $y_2 \ge 0$ and describes this as the default condition for the dual variables in this setup.

1. Stated for the dual variables in this example.

1. Dual objective coefficients come from primal right-hand sides
2. Dual constraints come from primal variable columns

1. Audio
   Observation

   The instructor says the RHS of the constraints is used for the objective function and the coefficients are used for writing down the constraint.

2. Diagram
   Observation

   The board shows a completed primal-to-dual example with the dual objective coefficients matching the primal RHS values and the dual constraint coefficients matching the primal column coefficients.

For this lecture, the dual is built by taking the right-hand-side constants of the primal constraints as the coefficients of the dual objective function, and by using the primal constraint coefficients to form the dual constraints.

1. The primal must first be in the required standard form before applying the rule.

1. Audio
   Observation

   The instructor states that if it is minimize, all the constraints should be greater than equal to.

2. Diagram
   Observation

   Problem 2 on the board is a minimization problem whose third constraint is written with $\le$.

In this video, a minimization primal is treated as ready for duality only when every constraint is written in the form $\ge$.

1. Applies to the minimization primal shown in Problem 2.

1. Audio
   Observation

   The instructor says the third constraint needs to be multiplied with the minus sign and converted into greater than equal to.

2. Formula
   Observation

   The board changes $2x_1 - x_2 \le 6$ into -2x_1 + $x_2 \ge -6$.

To bring $a \le$ constraint into the required minimization standard form, multiply the whole inequality by -1, which reverses the inequality direction and changes the signs of both sides.

1. Used when the primal is a minimization problem and one constraint is written as $\le$.

1. Standard form condition for a minimization primal

1. Audio
   Observation

   The instructor says "duality and dual simplex method, both are different" and later repeats that finding the dual is different from the dual simplex method.

In this lecture segment, the instructor claims that the topic of duality (primal and dual problems) is different from the dual simplex method.

1. The problem statement asks to find the dual of the given problems.

2. The instructor is contrasting terminology, not proving a theorem.

For the present lesson context.

1. Audio
   Observation

   The instructor says for finding the dual, the first step is to check whether the primal is in the correct form or not.

Before constructing the dual, one must first check whether the given primal problem is already in the required standard form.

1. The task is to form the dual of a given primal linear program.

For the procedure being taught in this clip.

1. Audio
   Observation

   The instructor says: Maximize, its dual will become minimize. If it is minimize, its dual will become maximize.

2. Formula
   Observation

   The note box on the board reads Max -> <= and Min -> >=, visually reinforcing the opposite-direction pairing.

In the rule system used in this clip, the dual of a maximization problem is a minimization problem, and the dual of a minimization problem is a maximization problem.

1. The problem is being converted according to the primal-dual rules shown on the board.

For the primal-dual pair discussed in this example.

1. Audio
   Observation

   The instructor says the problem is maximize and all the constraints are less than equal to, therefore the primal is in the correct form.

2. Formula
   Observation

   The displayed primal is Max $Z = 40x_1 + 35x_2$ subject to $2x_1 + 3x_2 \le 60$, $4x_1 + 8x_2 \le 96$, $x_1$,$x_2 \ge 0$.

The given primal problem is already in the correct form for direct dual construction because it is a maximization problem with $\le$ constraints and nonnegative variables.

1. The problem shown on the left side of the board is the primal under consideration.

For the specific displayed primal problem.

1. Audio
   Observation

   The instructor explicitly states the condition for the primal to be correct in the minimization case.

If the primal is a minimization problem, then all its constraints should be greater than or equal to before forming the dual.

1. The problem is a minimization linear program.

For all constraints in the primal minimization problem.

1. Audio
   Observation

   The instructor explains that the first step is to check whether the primal is in the correct form, then restates the max/min inequality rules while pointing at the board.

2. Formula
   Observation

   Problem 1 on the board is Max with $\le$ constraints; Problem 2 is Min with $\ge$ constraints.

1. Expression

   $$\text{Given primal problem}$$
   Explanation

   Start from the problem stated on the board, identified by the instructor as the primal.

   Justification

   The instructor says "primal means what? question" and points to the listed problems.

   Shown in the video
2. Expression

   $$\text{Check objective direction: Max or Min}$$
   Explanation

   Identify whether the primal is a maximization or minimization problem.

   Justification

   The board explicitly labels Problem 1 as Max and Problem 2 as Min.

   Shown in the video
3. Expression

   $$\text{If Max, require all constraints } \le$$
   Explanation

   For a maximization primal, inspect that every constraint uses $\le$.

   Justification

   This is the rule written as "Max -> <=" and spoken by the instructor.

   Shown in the video
4. Expression

   $$\text{If Min, require all constraints } \ge$$
   Explanation

   For a minimization primal, inspect that every constraint uses $\ge$.

   Justification

   This is the rule written as "Min -> >=" and spoken by the instructor.

   Shown in the video
5. Expression

   $$\text{Then proceed to form the dual}$$
   Explanation

   Only after confirming the primal is in the correct form does the lecture move toward constructing the dual.

   Justification

   The instructor explicitly calls this the first step for finding the dual.

   Shown in the video

The clip establishes a preliminary checklist for dual formation: identify the primal, verify the objective direction, and confirm the constraint inequalities match the required standard form.

1. Audio
   Observation

   The instructor narrates each step while writing: switch max to min, use RHS values 60 and 96 in the objective, use coefficients of $x_1$ and $x_2$ to form constraints, then add $y_1$,$y_2 \ge 0$.

2. Formula
   Observation

   The final dual written on the board is Min $W = 60y_1 + 96y_2$ subject to $2y_1 + 4y_2 \ge 40$, $3y_1 + 8y_2 \ge 35$, $y_1$,$y_2 \ge 0$.

1. Expression

   $$\text{Primal: Max } Z = 40x_1 + 35x_2 \text{ s.t. } 2x_1 + 3x_2 \le 60,\; 4x_1 + 8x_2 \le 96,\; x_1,x_2 \ge 0$$
   Explanation

   Start from the given primal problem on the board.

   Justification

   Directly visible formula evidence.

   Shown in the video
2. Expression

   $$\text{Dual objective sense: Min } W$$
   Explanation

   Because the primal is a maximization problem, the dual becomes a minimization problem.

   Justification

   Stated by the instructor as the rule for converting primal to dual.

   Shown in the video
3. Expression

   $$W = 60y_1 + 96y_2$$
   Explanation

   Use the right-hand sides of the two primal constraints as coefficients of the two dual variables.

   Justification

   The instructor explicitly says the objective function is formed from the right-hand side of the constraints.

   Shown in the video
4. Expression

   $$2y_1 + 4y_2 \ge 40$$
   Explanation

   Take the column of coefficients of $x_1$ from the primal constraints, namely 2 and 4, and set their combination against the primal objective coefficient 40.

   Justification

   The instructor says the constraints are formed based on the coefficient of $x_1$ and uses 40 from the objective function.

   Shown in the video
5. Expression

   $$3y_1 + 8y_2 \ge 35$$
   Explanation

   Take the column of coefficients of $x_2$ from the primal constraints, namely 3 and 8, and set their combination against the primal objective coefficient 35.

   Justification

   The instructor repeats the same rule for $x_2$ and uses 35 from the objective function.

   Shown in the video
6. Expression

   $$y_1, y_2 \ge 0$$
   Explanation

   Append nonnegativity restrictions for the dual variables.

   Justification

   The instructor calls this the default condition and writes it explicitly.

   Shown in the video

The dual of the displayed primal is Min $W = 60y_1 + 96y_2$ subject to $2y_1 + 4y_2 \ge 40$, $3y_1 + 8y_2 \ge 35$, $y_1$,$y_2 \ge 0$.

1. Audio
   Observation

   The instructor explains that the third constraint is problematic because the primal is minimize and therefore must be converted.

2. Formula
   Observation

   The rewritten line on the board is -2x_1 + $x_2 \ge -6$.

1. Expression

   $$2x_1 - x_2 \le 6$$
   Explanation

   Start from the third constraint as originally written in Problem 2.

   Justification

   Observed directly on the board.

   Shown in the video
2. Expression

   $$-2x_1 + x_2 \ge -6$$
   Explanation

   Multiply the entire inequality by -1 and reverse the inequality sign.

   Justification

   Stated verbally by the instructor and shown on the board.

   Shown in the video

The third constraint is converted into the required minimization standard form.

1. Audio
   Observation

   Instructor states: 'is 18, 8 and minus 6. So 18 y1 plus 8 y2 minus 6 y3.'

2. Formula
   Observation

   Whiteboard shows the primal RHS values 18, 8, -6 being used to form the dual objective function Max $W = 18y_1 + 8y_2 - 6y_3$.

1. Expression

   $$\text{Primal RHS constants: } 18, 8, -6$$
   Explanation

   Identify the right-hand side constants of the primal constraints.

   Justification

   Direct observation from the primal problem statement on the whiteboard.

   Shown in the video
2. Expression

   $$\max W = 18y_1 + 8y_2 - 6y_3$$
   Explanation

   Form the dual objective function by multiplying each primal RHS constant by its corresponding dual variable and summing them up, changing the optimization direction to maximization.

   Justification

   Standard rule for constructing the dual objective function from a primal minimization problem.

   Shown in the video

The dual objective function is correctly formulated as Max $W = 18y_1 + 8y_2 - 6y_3$.

1. Audio
   Observation

   Instructor states: 'subject to with x1 we have 3, here we have 1, here we have minus 2. So that is 3 y1 then we have 1 so that is y2... these are the coefficient of the first constraint that is minus 2 y3 and since it is maximize it will be converted into less than equal to less than equal to 10.'

2. Formula
   Observation

   Whiteboard shows the first column of the primal constraint matrix (3, 1, -2) being used to form the first dual constraint: $3y_1 + y_2 - 2y_3 \leq 10$.

1. Expression

   $$\text{Primal coefficients for } x_1: 3, 1, -2$$
   Explanation

   Identify the coefficients of the first primal variable $x_1$ across all three primal constraints.

   Justification

   Direct observation from the primal problem statement on the whiteboard.

   Shown in the video
2. Expression

   $$\text{Primal objective coefficient for } x_1: 10$$
   Explanation

   Identify the coefficient of $x_1$ in the primal objective function.

   Justification

   Direct observation from the primal objective function Min $Z = 10x_1 + 20x_2$.

   Shown in the video
3. Expression

   $$3y_1 + y_2 - 2y_3 \leq 10$$
   Explanation

   Form the first dual constraint by taking the dot product of the primal $x_1$ coefficients with the dual variables, setting it less than or equal to the primal objective coefficient for $x_1$.

   Justification

   Standard rule for constructing dual constraints from a primal minimization problem to a dual maximization problem.

   Shown in the video

The first dual constraint is correctly formulated as $3y_1 + y_2 - 2y_3 \leq 10$.

1. Audio
   Observation

   Instructor states: 'Next with x2 we have 2, 3 and 1. So 2 3 and 1 less than equal to with x2 in the objective function we have 20.'

2. Formula
   Observation

   Whiteboard shows the second column of the primal constraint matrix (2, 3, 1) being used to form the second dual constraint: $2y_1 + 3y_2 + y_3 \leq 20$.

1. Expression

   $$\text{Primal coefficients for } x_2: 2, 3, 1$$
   Explanation

   Identify the coefficients of the second primal variable $x_2$ across all three primal constraints.

   Justification

   Direct observation from the primal problem statement on the whiteboard.

   Shown in the video
2. Expression

   $$\text{Primal objective coefficient for } x_2: 20$$
   Explanation

   Identify the coefficient of $x_2$ in the primal objective function.

   Justification

   Direct observation from the primal objective function Min $Z = 10x_1 + 20x_2$.

   Shown in the video
3. Expression

   $$2y_1 + 3y_2 + y_3 \leq 20$$
   Explanation

   Form the second dual constraint by taking the dot product of the primal $x_2$ coefficients with the dual variables, setting it less than or equal to the primal objective coefficient for $x_2$.

   Justification

   Standard rule for constructing dual constraints from a primal minimization problem to a dual maximization problem.

   Shown in the video

The second dual constraint is correctly formulated as $2y_1 + 3y_2 + y_3 \leq 20$.

1. Formula
   Observation

   Left side of the board shows: Max $Z = 40x_1 + 35x_2$; Sub to $2x_1 + 3x_2$ <= 60; $4x_1 + 3x_2$ <= 96; $x_1$, $x_2$ >= 0.

2. Audio
   Observation

   The instructor refers to two problems given in front of us and asks to find the dual of the following problems.

Find the dual of the following problem: Maximize $Z = 40x_1 + 35x_2$ subject to $2x_1 + 3x_2 \le 60$, $4x_1 + 3x_2 \le 96$, and $x_1, x_2 \ge 0$.

1. Objective is maximization.

2. Two decision variables $x_1,x_2$.

3. Two resource-type constraints with right-hand sides 60 and 96.

4. Nonnegativity conditions $x_1,x_2\ge 0$.

Use this as a primal problem for the dual-formation discussion.

1. Expression

   $$\text{Max } Z = 40x_1 + 35x_2$$
   Explanation

   Read the objective function from the board.

   Justification

   Directly visible in the left-column example.

   Shown in the video
2. Expression

   $$2x_1 + 3x_2 \le 60,\quad 4x_1 + 3x_2 \le 96$$
   Explanation

   Read the two constraints.

   Justification

   Directly visible on the board.

   Shown in the video
3. Expression

   $$x_1, x_2 \ge 0$$
   Explanation

   Read the variable restrictions.

   Justification

   Directly visible on the board.

   Shown in the video
4. Expression

   $$\text{This is a Max problem with all } \le \text{ constraints}$$
   Explanation

   Classify the example against the rule box.

   Justification

   Matches the spoken and written rule "Max -> <=".

   Derived from the video

The clip presents this as a primal maximization problem already in the stated standard inequality form.

The classification is checked against the board rule "Max -> <=", and all displayed constraints are indeed $\le$.

1. Formula
   Observation

   Right side of the board shows: Min $Z = 10x_1 + 20x_2$; Sub to $3x_1 + 2x_2$ >= 18; $x_1 + 3x_2$ >= 8; $2x_1 - x_2$ <= 6; $x_1$, $x_2$ >= 0.

2. Audio
   Observation

   The instructor again refers to the listed problems and the need to check whether the primal is in correct form.

Find the dual of the following problem: Minimize $Z = 10x_1 + 20x_2$ subject to $3x_1 + 2x_2 \ge 18$, $x_1 + 3x_2 \ge 8$, $2x_1 - x_2 \le 6$, and $x_1, x_2 \ge 0$.

1. Objective is minimization.

2. Two decision variables $x_1,x_2$.

3. Three constraints, two of type $\ge$ and one of type $\le$.

4. Nonnegativity conditions $x_1,x_2\ge 0$.

Use this as a second primal example for the standard-form check discussed in the clip.

1. Expression

   $$\text{Min } Z = 10x_1 + 20x_2$$
   Explanation

   Read the objective function from the board.

   Justification

   Directly visible in the right-column example.

   Shown in the video
2. Expression

   $$3x_1 + 2x_2 \ge 18,\quad x_1 + 3x_2 \ge 8,\quad 2x_1 - x_2 \le 6$$
   Explanation

   Read the three constraints.

   Justification

   Directly visible on the board.

   Shown in the video
3. Expression

   $$x_1, x_2 \ge 0$$
   Explanation

   Read the variable restrictions.

   Justification

   Directly visible on the board.

   Shown in the video
4. Expression

   $$\text{This Min problem is not uniformly } \ge$$
   Explanation

   Compare the displayed constraints with the rule for minimization problems.

   Justification

   The instructor's rule says Min problems should have all constraints $\ge$, but the third displayed constraint is $\le$.

   Derived from the video

The clip shows this as a primal minimization example whose displayed constraints do not all match the stated minimization standard form.

Checked against the board rule "Min -> >=": two constraints satisfy it, but $2x_1 - x_2 \le 6$ does not.

1. Formula
   Observation

   The board heading says Find the dual of the following problems, and problem 1 is fully solved on the left.

2. Audio
   Observation

   The instructor concludes: So this is the primal dual of the above primal.

Find the dual of Max $Z = 40x_1 + 35x_2$ subject to $2x_1 + 3x_2 \le 60$, $4x_1 + 8x_2 \le 96$, $x_1$,$x_2 \ge 0$.

1. Primal objective: Max $Z = 40x_1 + 35x_2$.

2. First primal constraint: $2x_1 + 3x_2 \le 60$.

3. Second primal constraint: $4x_1 + 8x_2 \le 96$.

4. Nonnegativity: $x_1$,$x_2 \ge 0$.

Write the corresponding dual linear program.

1. Expression

   $$\text{Check form: Max with } \le \text{ constraints and } x_1,x_2 \ge 0$$
   Explanation

   Confirm the primal is already in the required form before constructing the dual.

   Justification

   Explicitly stated by the instructor.

   Shown in the video
2. Expression

   $$\text{Min } W = 60y_1 + 96y_2$$
   Explanation

   Replace the primal maximization by a dual minimization and use the primal right-hand sides 60 and 96 as objective coefficients.

   Justification

   Rule stated in the audio and written on the board.

   Shown in the video
3. Expression

   $$2y_1 + 4y_2 \ge 40$$
   Explanation

   Form the first dual constraint from the $x_1$ column of the primal and the primal objective coefficient 40.

   Justification

   The instructor says the constraints are based on the coefficient of $x_1$.

   Shown in the video
4. Expression

   $$3y_1 + 8y_2 \ge 35$$
   Explanation

   Form the second dual constraint from the $x_2$ column of the primal and the primal objective coefficient 35.

   Justification

   The instructor repeats the same rule for $x_2$.

   Shown in the video
5. Expression

   $$y_1, y_2 \ge 0$$
   Explanation

   Add the default nonnegativity restrictions for the dual variables.

   Justification

   Written and verbally identified as default by the instructor.

   Shown in the video

Min $W = 60y_1 + 96y_2$ subject to $2y_1 + 4y_2 \ge 40$, $3y_1 + 8y_2 \ge 35$, $y_1$,$y_2 \ge 0$.

The answer matches the final handwritten dual on the board and follows the rule sequence stated aloud by the instructor.

1. Audio
   Observation

   The instructor introduces problem number two and checks whether the primal is in the correct form.

2. Formula
   Observation

   The board shows Min $Z = 10x_1 + 20x_2$ with constraints $3x_1 + 2x_2 \ge 18$, $x_1 + 3x_2 \ge 8$, $2x_1 - x_2 \le 6$, $x_1$, $x_2 \ge 0$.

3. Formula
   Observation

   The rewritten primal on the board becomes Min $Z = 10x_1 + 20x_2$ subject to $3x_1 + 2x_2 \ge 18$, $x_1 + 3x_2 \ge 8$, -2x_1 + $x_2 \ge -6$, $x_1$, $x_2 \ge 0$.

Given the minimization problem Min $Z = 10x_1 + 20x_2$ subject to $3x_1 + 2x_2 \ge 18$, $x_1 + 3x_2 \ge 8$, $2x_1 - x_2 \le 6$, $x_1$, $x_2 \ge 0$, rewrite the primal in the required standard form before finding the dual.

1. Objective: Min $Z = 10x_1 + 20x_2$

2. Constraint 1: $3x_1 + 2x_2 \ge 18$

3. Constraint 2: $x_1 + 3x_2 \ge 8$

4. Constraint 3: $2x_1 - x_2 \le 6$

5. Non-negativity: $x_1$, $x_2 \ge 0$

Convert the primal into the standard form needed for duality.

1. Expression

   $$3x_1 + 2x_2 \ge 18$$
   Explanation

   Keep the first constraint unchanged because it already satisfies the minimization standard form.

   Justification

   Stated by the instructor as perfect and kept as it is.

   Shown in the video
2. Expression

   $$x_1 + 3x_2 \ge 8$$
   Explanation

   Keep the second constraint unchanged because it also satisfies the minimization standard form.

   Justification

   Stated by the instructor as keep it as it is.

   Shown in the video
3. Expression

   $$-2x_1 + x_2 \ge -6$$
   Explanation

   Replace the third constraint by multiplying through by -1 so that the inequality becomes $\ge$.

   Justification

   Explicitly explained and written on the board.

   Shown in the video
4. Expression

   $$x_1, x_2 \ge 0$$
   Explanation

   Retain the non-negativity restrictions.

   Justification

   Written on the board in the rewritten primal.

   Shown in the video

The prepared primal is Min $Z = 10x_1 + 20x_2$ subject to $3x_1 + 2x_2 \ge 18$, $x_1 + 3x_2 \ge 8$, -2x_1 + $x_2 \ge -6$, $x_1$, $x_2 \ge 0$.

All constraints are now in $\ge$ form, matching the instructor’s stated condition for a minimization primal.

1. Formula
   Observation

   The left side of the board shows Max $Z = 40x_1 + 35x_2$ with constraints $2x_1 + 3x_2 \le 60$, $4x_1 + 3x_2 \le 90$, $x_1$, $x_2 \ge 0$.

2. Formula
   Observation

   Below it the board shows Dual is given by Min $W = 60y_1 + 90y_2$ with constraints $2y_1 + 4y_2 \ge 40$, $3y_1 + 3y_2 \ge 35$, $y_1$, $y_2 \ge 0$.

3. Audio
   Observation

   The instructor summarizes how the dual is written from the primal data.

Use the displayed primal maximization example to identify how the dual is constructed.

1. Primal: Max $Z = 40x_1 + 35x_2$

2. Constraints: $2x_1 + 3x_2 \le 60$, $4x_1 + 3x_2 \le 90$

3. Non-negativity: $x_1$, $x_2 \ge 0$

Read off the dual construction pattern from the completed example.

1. Expression

   $$Min W = 60y_1 + 90y_2$$
   Explanation

   The dual objective coefficients 60 and 90 come from the primal constraint right-hand sides.

   Justification

   Visible on the board and summarized verbally by the instructor.

   Shown in the video
2. Expression

   $$2y_1 + 4y_2 \ge 40$$
   Explanation

   The first dual constraint uses the coefficients of $x_1$ from the two primal constraints and the primal objective coefficient of $x_1$ on the right-hand side.

   Justification

   Visible on the board and consistent with the instructor’s explanation.

   Shown in the video
3. Expression

   $$3y_1 + 3y_2 \ge 35$$
   Explanation

   The second dual constraint uses the coefficients of $x_2$ from the two primal constraints and the primal objective coefficient of $x_2$ on the right-hand side.

   Justification

   Visible on the board and consistent with the instructor’s explanation.

   Shown in the video

The displayed dual is Min $W = 60y_1 + 90y_2$ subject to $2y_1 + 4y_2 \ge 40$, $3y_1 + 3y_2 \ge 35$, $y_1$, $y_2 \ge 0$.

The board itself presents this as the completed dual of the first example.

1. Formula
   Observation

   The entire segment focuses on converting a specific primal minimization problem into its dual maximization problem on the whiteboard.

Given the primal linear programming problem: Min $Z = 10x_1 + 20x_2$ subject to $3x_1 + 2x_2 \geq 18$, $x_1 + 3x_2 \geq 8$, -2x_1 + $x_2 \geq -6$, and $x_1$, $x_2 \geq 0$. Find its dual problem.