Lagrange Multiplier
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Example 1:
Question: Minimize the function f(x,y) = subject to the constrain 3x-y+1 = 0.
Solution:
- 1.) Use Lagrange Multiplier.
- 2.) Solve for y in the constrain.
- 3.) Minimize by using the constrain g(x,y) = 3x-y+1 = 0
- 4.) Let F(x,y,λ) = f(x,y) + λ(g(x,y)) = Failed to parse (syntax error): {\displaystyle (1/2)x^2-3xy+y^2+λ(3x-y+1)}
- 5.) F(x,y,λ) = Failed to parse (syntax error): {\displaystyle (1/2)x^2-3xy+y^2+3λx-λy+λ)}
- ∂F/∂x = x -3y + 3λ = 0
- ∂F/∂y = -3x + 2y - λ = 0
- ∂F/∂λ = 3x - y + 1 = 0
- x - 3y = -3λ
- x-3y/-3 = λ
- -3x + 2y = λ
- x-3y/-3 = -3x + 2y
- x-3y = 9x - 6y
- 0 = 8x - 3y
- 3x + 1 = y
- 0 = 8x - 3(3x+1)
- 0 = 8x - 9x -3
- 0 = -x - 3
- x = -3
- Therefore (-3, -8) is the correct answer.
Example 2:
Question: Suppose that a units of labor and b units of capital can produce f(a,b) = 60a^{3/4}b^{1/4} units of the product. Each unit of capital costs $200. Each unit of labor costs $100. The limit spent on production is $30000. What is the number of units of labor and capital that should be used to maximize production?
Solution:
- Constraint equation is: 100a + 200b = 30000
- In other words, it is: h(a,b) = 30000 - 100x - 200y = 0.
- The function we need to maximize is f(x,y) = 60a^{3/4}b^{1/4}.
- The Lagrange Multiplier is: F(a,b,λ) = 60a^{3/4}b^{1/4} + λ(30000 - 100x - 200y)^{1/4}
- ∂F/∂a = 45a^{-1/4}b^{1/4} - 100λ = 0
- ∂F/∂b = 15a^{3/4})b^{-3/4} - 200λ = 0
- ∂F/∂λ = 30000 - 100a - 200b = 0
- Solve for λ for the first two equations and get:
- λ = 45/100a^{-1/4}b^{1/4} = 9/20a^{1/4}b^{1/4}
- λ = 15/200a^{3/4}b^{-3/4} = 3/40a^{3/4}b^{-3/4}
- 9/20a^{-1/4}b^{1/4} = 3/40a^{3/4}b^{-3/4}
- 9/20b = 3/40a
- y = 1/6a
- 100a + 200(1/6a) = 30000
- 400/3a = 30000
- a = 225
- b = 225/6 = 37.5
- Solve for λ for the first two equations and get:
- The number of units of labor is 225 and the number of units of capital is 37.5 to obtain maximum production.
Example 3:
Question: Minimize 42x + 28y, with constraint 600 - xy = 0 ( x and y are restricted to positive values.)
Solution:
- Constraint equation is: 600 - xy = 0
- In other words, it is: h(x,y) = 600 - xy = 0.
- The function we need to minimize is f(x,y) = 42x + 28y.
- The Lagrange Multiplier is: F(x,y,λ) = 42x + 28y + λ(600 - xy).
- ∂F/∂x = 42 - λy = 0
- ∂F/∂y = 28 - λx = 0
- ∂F/∂λ = 600 - xy = 0
- Solve for λ for the first two equations and get:
- λ = 42/y = 28/x
- 42x = 28y
- x = 2/3y
- 600 - (2/3y)y = 0
- y^{2}= 3/2(600) = 900
- y = +/- 30
- y = -30 is not going to be used because we are only using positive values of x and y.
- y = 30
- x = 2/3(30) = 20
- λ = 28/20 = 7/5
- Solve for λ for the first two equations and get:
- Minimum value is obtained when x = 20 and y = 30 and λ = 7/5.
- The minimum value is 42(20) + 28(30) = 1680.
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