Lagrange Multiplier
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Example 1:
Question: Minimize the function f(x,y) = subject to the constrain 3xy+1 = 0.
Solution:
 1.) Use Lagrange Multiplier.
 2.) Solve for y in the constrain.
 3.) Minimize by using the constrain g(x,y) = 3xy+1 = 0
 4.) Let F(x,y,λ) = f(x,y) + λ(g(x,y)) =
 5.) F(x,y,λ) =
 ∂F/∂x = x 3y + 3λ = 0
 ∂F/∂y = 3x + 2y  λ = 0
 ∂F/∂λ = 3x  y + 1 = 0
 x  3y = 3λ
 x3y/3 = λ
 3x + 2y = λ
 x3y/3 = 3x + 2y
 x3y = 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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