Chapter 12 - lagrange multipliers study guide - MA129 Week 12 Lagrange Multipliers Antiderivatives 1 Lagrange Multipliers(9.4 Method For Finding

# Chapter 12 - lagrange multipliers study guide - MA129 Week...

• Test Prep
• 2

This preview shows page 1 - 2 out of 2 pages.

MA129 Week 12 - Lagrange Multipliers; Antiderivatives 1. Lagrange Multipliers (9.4) Method For Finding Relative Extrema Subject To A Constraint Consider the functions f ( x, y ) and g ( x, y ) . To find any relative extrema of f ( x, y ) subject to the constraint g ( x, y ) = 0 (i.e. relative max/min of f ( x, y ) that lie on g ( x, y ) = 0): Step 1: Express constraint in the form g ( x, y ) = 0 . Step 2: Determine the Lagrange function F ( x, y, λ ) = f ( x, y ) - λg ( x, y ) . Step 3: Find F x ( x, y, λ ) , F y ( x, y, λ ) , and F λ ( x, y, λ ) . Step 4: Solve: F x ( x, y, λ ) = 0 F y ( x, y, λ ) = 0 F λ ( x, y, λ ) = 0 Step 5: Solution will involve relative extrema of f ( x, y ) subject to g ( x, y ) = 0 . Example: Find the relative maximum of f ( x, y ) = x 2 - 10 y 2 subject to the constraint x - y = 18 . Step 1: f ( x, y ) = x 2 - 10 y 2 and g ( x, y ) = x - y - 18 = 0 Step 2: F ( x, y, λ ) = f ( x, y ) - λg ( x, y ) = x 2 - 10 y 2 - λ ( x - y - 18) = x 2 - 10 y 2 - λx + λy + 18 λ Step 3: F x ( x, y, λ ) = 2 x - λ F y ( x, y, λ ) = - 20 y + λ F λ ( x, y, λ ) = - x + y + 18 Step 4: Solve: 2 x - λ = 0 2 x = λ (1) - 20 y + λ = 0 20 y = λ (2) - x + y + 18 = 0 - x + y + 18 = 0 (3) From (1) and (2), 2 x = 20 y x = 10 y (4) Substituting (4) into (3): - (10 y ) + y + 18 = 0 ⇒ - 9 y + 18 = 0 y = 2 Substituting y = 2 into (4): x = 10(2) = 20 Step 5: Relative maximum of f ( x, y ) subject to x - y = 18 is f (20 , 2) = 360 . 1
2. Indefinite Integrals and Antiderivatives (7.1) A function F is called an antiderivative of the continuous function
• • • 