Lagrange Multiplier Analysis on Non-Linear Constraints: Economic Modeling with Complex Constraints

Exploration of Lagrange's Concept in Financial Portfolio Optimization: Finding the Best Investment with Constraints Name: Alvinska Hebat Grade: 910-Math LAGRANGE FUNCTION Lagrange Function: Function Optimization with Constraints • • • Lagrange function is a function that is formed to optimize the function we are aiming at (objective function) by including its control function. The optimal values obtained from the process of optimizing the objective function through the formation of the Lagrange function are simultaneously limited by the constraints that accompany the objective function. If we have an objective function to be optimized: Y = f(X,Z) with constraints: a - m1X m2 Z = 0 then the Lagrange function (V) is: objective function plus λ multiplied by the constraint function λ = Lagrange coefficient. Thus the Lagrange function is: V = f(X,Z) + λ(a - m1X - m2 Z) To optimize Y = f(X,Z) with the constraint function: a - m1X - m2 Z = 0, there are 3 (three) conditions that must be met, namely: Example: Optimize Y = 10X2Z with constraint 900 = 2X +3Z Answer: To optimize Y = 10X2Z with the constraint 900 = 2X + 3Z, the Lagrange function (V) must first be formed as follows: V = 10X2Z + λ (900 -2X - 3Z) The optimization process must meet the following conditions: Eliminate λ from equations (1) and (2): (1) x 3 → 60XZ - 6 λ = 0 (2) x 2 → 20X2 - 6 λ = 0 - 60XZ - 20X2 = 0 60XZ = 20X2 3Z = X → X = 3Z Substitute X = 3Z into equation (3) Equation (3) → 0 = 900 - 2X - 3Z 0 = 900 - 2(3Z) - 3Z 0 = 900 - 6Z - 3Z 0 = 900 - 9Z or 900 = 9Z Then Z = 900/9 = 100 So that X = 3Z = 3(100) = 300 After finding the optimal X and Z values, we can find the optimal Y value as follows: Y = 10X2Z Y = 10(300)2 (100) = 90,000,000 So the optimal Y condition can occur with the value of X = 300 and Z = 100 With the optimal Y value = 90,000,000 Application of Lagrange Function in Economic Concepts • Lagrange functions for economic concepts are applied in certain functions that must be optimized with accompanying constraints. • • For example, the goal of optimizing production levels with budget constraints that producers have to buy inputs used for the production process, or conversely the goal of optimizing input costs with the constraints of their production levels. The Lagrange function can also be used to optimize the utility (satisfaction) of consumers in consuming goods through the budget constraints that consumers have, or vice versa to optimize the consumer budget for goods consumed at a certain level of utility. Example: Through the Lagrange function method maximize the production level of good X in the relationship: Q = f(K,L) = 150 K 0.3L 0.7 with the following constraints: 30,000 = 3K + 1.5L Description: Q = total output in kilograms; K = total capital goods; L = total labor hours. Ask: what is the amount of capital goods and labor hours needed to maximize output? What is also the magnitude of the maximum output? Answer: Lagrange function (V): V = 150 K 0.3L 0.7 + λ(30,000 - 3K - 1.5L) = 150 K 0.3L 0.7 + 30,000 λ - 3K λ - 1.5 L λ The optimization process requires the following conditions: (1) dV/dK = 0 = 45K -0.7L 0.7 - 3 λ (2) dV/dL = 0 = 105K 0.3L -0.3 - 1.5 λ (3) dV/d λ = 0 = 30,000 - 3K - 1.5L From equations (1) and (2) eliminate λ: (1) x 1 → 45K-0.7L 0.7 - 3 λ = 0 (2) x 2 → 210K0.3L -0.3 - 3 λ = 0 - 45K-0.7L 0.7 - 210K0.3L -0.3 = 0 45K-0.7L 0.7 = 210K0.3L -0.3 45L0.3L 0.7 = 210 K0.3K 0.7 45L = 210K L = (210/45)K → L = 4.667K Substitute L = 4.667K into equation (3): (3)→ 30,000 - 3K - 1.5L = 0 30,000 - 3K 1.5(4.667K) = 0 30,000 - 3K - 7K = 0 30,000 - 10K = 0 30,000 = 10K → hence K = 30,000/10 = 3000 Since K = 3000 then L = 4.667K = 4.667 (3000) = 14,000 so that the optimal Q is : Q = 150 K 0.3L 0.7 = 150 (3,000)0.3(14,000)0.7 = 1,322,870.283 Conclusion: the optimal production level of output X of 1,322,870.283 kilograms is generated by the use of capital goods in the amount of 3,000 units and 14,000 labor hours.