Lagrange multiplier example, part 2

Khan Academy
15 Nov 201606:50
EducationalLearning
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TLDRIn this instructional video, the instructor tackles an optimization problem involving three unknowns: tons of steel (s), hours of labor (h), and a Lagrange Multiplier (lambda). Through simplification and substitution, the instructor derives equations to solve for s and h, ultimately finding the optimal allocation of resources for a budget of $20,000. The solution involves buying 10/3 tons of steel and 2,000/3 hours of labor, maximizing revenue for a widget production model with given costs.

Takeaways
  • ๐Ÿ” The script discusses solving a system of equations with three unknowns: tons of steel (s), hours of labor (h), and the Lagrange Multiplier (lambda).
  • ๐Ÿ“ The instructor introduces a substitution method using 'u' for s/h to simplify the equations and make the process more manageable.
  • ๐Ÿ“š The equations are derived from a constraint and a revenue function, aiming to find the optimal allocation of resources within a given budget.
  • ๐Ÿ”ข The instructor simplifies the equations by combining constants and using algebraic manipulation to isolate 'u' and lambda.
  • ๐Ÿงฉ The process involves multiplying each equation by a form of 'u' to the appropriate power to make the equations more comparable and easier to solve.
  • โš–๏ธ The equations are then balanced by multiplying one by specific factors to match the other side, leading to a common expression involving 'u'.
  • ๐Ÿ“‰ The result of the manipulation is an equation that equates 200 times 'u' to a constant, which simplifies to a direct relationship between 's' and 'h'.
  • ๐Ÿ”„ The substitution of 'h' with 200 times 's' is applied to the constraint equation to find the values of 's' and 'h'.
  • ๐Ÿ’ฐ The budget constraint is used to solve for 's', the tons of steel, which is found to be 10/3 tons.
  • โฑ Using the relationship h = 200 * s, the hours of labor (h) are calculated to be 2000/3 hours.
  • ๐Ÿ“Š The final takeaway is the optimal allocation for the given problem: buy 10/3 tons of steel and use 2000/3 hours of labor to maximize revenue within a $20,000 budget.
Q & A

  • What are the three unknowns in the equations discussed in the script?

    -The three unknowns are s (tons of steel), h (hours of labor), and lambda (the Lagrange Multiplier).

  • What is the role of the Lagrange Multiplier in this context?

    -The Lagrange Multiplier, lambda, is used to find the optimal solution that satisfies the constraint by being a proportionality constant between the gradient vectors of the revenue function and the constraint function.

  • What substitution is made to simplify the equations?

    -The substitution made is to introduce a new variable u, where u is defined as s divided by h.

  • How is the first equation rewritten using the new variable u?

    -The first equation is rewritten as 200 thirds times u to the power of one third equals 20 times lambda.

  • What happens when the second equation is rewritten using u?

    -The second equation becomes 100 thirds times u to the negative two thirds, due to the swapping of h and s.

  • What is the purpose of multiplying the equations by u to the two thirds?

    -Multiplying by u to the two thirds simplifies the equations by turning the left side into just u, and making the right sides similar, which helps in isolating u.

  • What does the equation 200 times u equal to 60 times lambda times u to the two thirds imply?

    -This implies that 200 times u is equal to 60 times lambda, as the u terms on both sides of the equation cancel out.

  • What is the simplified relationship between h and s derived from the equations?

    -The simplified relationship is h equals 200 times s.

  • How is the budget constraint equation used to find the values of h and s?

    -By substituting h with 200 times s into the budget constraint equation, the values of s and subsequently h are found by solving the resulting equation.

  • What is the final allocation of steel and labor hours according to the script?

    -The final allocation is 10/3 tons of steel and 2000/3 hours of labor.

Outlines
00:00
๐Ÿงฎ Simplifying the Equations

The instructor discusses simplifying equations to solve for the three unknowns: s (tons of steel), h (hours of labor), and ฮป (Lagrange multiplier). They introduce a substitution (u = s/h) to make the equations cleaner and then simplify the first equation to 200/3 u^(1/3) = 20ฮป and the second equation to 100/3 u^(-2/3) = 20ฮป.

05:02
๐Ÿ”„ Adjusting and Equalizing the Equations

The instructor further simplifies the equations by combining constants and isolating terms involving ฮป. They multiply the equations to eliminate the exponents on u, leading to simpler forms: 3/10ฮป u^(2/3) = u and 60ฮป u^(2/3) = 1. They equate the simplified forms to solve for u, resulting in the equation 200s/h = 1, and derive that h = 200s.

๐Ÿ”— Applying the Constraint Equation

The instructor substitutes h = 200s into the budget constraint equation 20h + 2000s = 20000. This substitution simplifies the equation to 6000s = 20000, solving for s = 10/3. Subsequently, they calculate h as 200 times 10/3, resulting in h = 2000/3. Finally, the optimal allocation is determined: 10/3 tons of steel and 2000/3 hours of labor.

Mindmap
Keywords
๐Ÿ’กLagrange Multiplier
A Lagrange Multiplier is a value used to find the local maxima and minima of a function subject to equality constraints. In the video, it is introduced as a proportionality constant, denoted by lambda, between the gradient vectors of the revenue function and the constraint function. This helps in optimizing the revenue given certain constraints.
๐Ÿ’กGradient Vectors
Gradient Vectors are vectors consisting of partial derivatives of a function. They point in the direction of the greatest rate of increase of the function. In the video, the gradient vectors of the revenue function and the constraint function are related through the Lagrange Multiplier, showing the optimization path.
๐Ÿ’กRevenue Function
The Revenue Function represents the total revenue generated from producing a certain number of goods. In the video, it's part of the optimization problem, where maximizing the revenue involves balancing the costs of steel and labor within a given budget.
๐Ÿ’กConstraint Function
A Constraint Function limits the values that the variables in an optimization problem can take. In the video, it represents the budget constraint of $20,000, which includes the costs of labor and steel.
๐Ÿ’กSubstitution
Substitution is a mathematical technique where one variable is replaced with another expression to simplify the equations. In the video, the instructor substitutes u for s divided by h to make the equations easier to solve.
๐Ÿ’กSimplification
Simplification involves reducing an equation or expression to a simpler form. The instructor simplifies the equations by combining constants and substituting variables to make solving for the unknowns more straightforward.
๐Ÿ’กOptimization
Optimization is the process of making something as effective or functional as possible. In the video, the goal is to optimize the allocation of steel and labor to maximize revenue within the given budget.
๐Ÿ’กBudget Constraint
A Budget Constraint is a limitation on the spending based on the available budget. In the video, the budget constraint of $20,000 limits the amount of steel and labor that can be purchased, guiding the optimization process.
๐Ÿ’กu to the power of one third
This mathematical expression represents the cube root of u. In the video, the instructor uses it as part of the simplification process to solve the equations for the unknowns s and h.
๐Ÿ’กs over h
The ratio of steel to labor, denoted as s over h, is a key variable in the optimization problem. The instructor introduces this ratio as u to simplify the problem and eventually solve for the optimal amounts of steel and labor.
Highlights

Introduction of two different equations with three unknowns: s (tons of steel), h (hours of labor), and lambda (Lagrange Multiplier).

Explanation of lambda as a proportionality constant between the gradient vectors of the revenue function and the constraint function.

The third equation is the constraint itself, aiding in solving for h, s, and lambda.

Simplification strategy by making a substitution, u for s divided by h, to ease the solving process.

Rewriting the first equation using the substitution u, simplifying the relationship between s and h.

Adjusting the second equation to reflect the inverse relationship of s and h, using u to the negative two thirds.

Combining constants and manipulating the equations to isolate u and lambda.

Multiplying both sides of the equations by specific factors to simplify and find a relationship between u and lambda.

Isolating u by multiplying each equation by u to the two thirds, simplifying the expressions.

Achieving a common right side for both equations, setting the stage for further simplification.

Multiplying the top equation by 200 to match the right side of the bottom equation, creating a direct comparison.

Deriving the equation 200 times u equals one, leading to the conclusion that h equals 200 times s.

Applying the relationship h = 200s to the constraint equation to find the values of h and s.

Solving the constraint equation to find the optimal amount of steel, s, as 10/3 tons.

Calculating the optimal hours of labor, h, as 2000/3, based on the derived relationship with s.

Conclusion of the optimal allocation for the revenue function model with given labor and steel costs and budget.

Transcripts

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OptimizationLagrange MultipliersMath TutorialProblem SolvingCalculusEconomicsEngineeringRevenue FunctionConstraint EquationMathematics