Lagrange Multipliers Example

Business Calculus · Axiom Academy

EXAMPLE Lagrange Multiplier Examples Step-by-step solutions to constrained optimization problems A consumer has utility function U(x, y) = xy and a budget of 120. Good x costs 2 per unit and good y costs 3 per unit. Find the consumption bundle that maximizes utility. Step 2: Find partial derivatives and set to zero Maximum utility: U(30, 20) = 600 We found λ = 10. This means: if the budget increased by 1 (from 120 to 121), the maximum utility would increase by approximately 10 units. λ is the marginal utility of income! A firm has production function Q = 4L^0.5 K^0.5. If the wage rate is w = 16 and the rental rate of capital is r = 4, find the cheapest way to produce Q = 40 units. Minimize cost C = 16L + 4K subject to 4L^0.5 K^0.5 = 40 Substitute K = 4L into constraint: Minimum cost: C = 16(25) + 4(100) = 800 The optimal K/L ratio is 4:1. Because capital is cheaper relative to labor (r/w = 4/16 = 0.25), the firm uses more capital than labor. The Lagrange multiplier λ = 8 represents the marginal cost of production—the cost of producing one more unit of output. 3 Profit Maximization with Resource Constraint A company makes two products. Profit from product A is 5x² and from product B is 8y. They share a combined resource constraint: x² + y² = 25. Maximize total profit. Maximum profit: P = 5(1) + 8(√24) ≈ 44.2 λ = 4/√24 ≈ 0.82 is the shadow price of the resource. If the resource constraint increased from 25 to 26, profit would increase by approximately 0.82.

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