Chapter 8Multivariable Calculus Section 4 Maxima and Minima Using Lagrange Multipliers
Learning Objectives for Section 8.4: Max/Min with Lagrange Multipliers • The student will be able to solve problems involving • Lagrange multipliers for functions of two independent variables • Lagrange multipliers for functions of three independent variables.
Theorem 1 - Lagrange Multipliers Step 1. Formulate the problem: Maximize (or minimize) z= f (x, y) subject to g (x, y) = 0. Step 2. Form the function F: Step 3. Find the critical points for F. That is, solve the system Step 4. Conclusion: If (x0, y0, λ0) is the only critical point of F, we assume that (x0, y0) is the solution. If F has more than one critical point, we evaluate z = f (x, y) at each of them, to determine the maximum or minimum.
Example Maximize f (x, y) = 25 – x2 – y2, subject to 2x + y = 10. Step 1. Formulate: Maximize z = f (x, y) = 25 – x2 – y2 subject to g(x, y) = 2x + y – 10 = 0 Step 2. Form the function F:
Example Maximize f (x, y) = 25 – x2 – y2, subject to 2x + y = 10. Step 1. Formulate: Maximize z = f (x, y) = 25 – x2 – y2 subject to g(x, y) = 2x + y – 10= 0 Step 2. Form the function F: Step 3. Find the critical points for F:
Example Maximize f (x, y) = 25 – x2 – y2, subject to 2x + y = 10. Step 1. Formulate: Maximize z = f (x, y) = 25 – x2 – y2 subject to g(x, y) = 2x + y – 10= 0 Step 2. Form the function F: Step 3. Find the critical points for F: Solving simultaneously yields one critical point at (4, 2, 4). Step 4. Conclusion:
Example Maximize f (x, y) = 25 – x2 – y2, subject to 2x + y = 10. Step 1. Formulate: Maximize z = f (x, y) = 25 – x2 – y2 subject to g(x, y) = 2x + y – 10= 0 Step 2. Form the function F: Step 3. Find the critical points for F: Solving simultaneously yields one critical point at (4, 2, 4). Step 4. Conclusion: Since (4, 2, 4) is the only critical point for F: Max of f (x, y) with constraints = f (4, 2) = 25 – 42 – 22 = 5.
Example 2 The Cobb-Douglas production function for a product is given by N(x, y) = 10 x0.6y0.4. Maximize N under the constraint that 30x + 60y = 300,000. Step 1. Formulate: Maximize N(x, y) = 10 x0.6y0.4subject to g(x, y) = 30x + 60y – 300,000 = 0 Step 2. Form the function F:
Example 2 The Cobb-Douglas production function for a product is given by N(x, y) = 10 x0.6y0.4. Maximize N under the constraint that 30x + 60y = 300,000. Step 1. Formulate: Maximize N(x, y) = 10 x0.6y0.4subject to g(x, y) = 30x + 60y – 300,000 = 0 Step 2. Form the function F:
Example 2(continued) Step 3. Find the critical points for F:
Example 2(continued) Step 3. Find the critical points for F: Solving yields one critical point (6000, 2000, – 0.1289) Step 4. Conclusion:
Example 2(continued) Step 3. Find the critical points for F: Solving yields one critical point (6000, 2000, – 0.1289) Step 4. Conclusion: Since (6000, 2000, – 0.1289) is the only critical point for F, we conclude that Max of N(x, y) under the given constraints = N(6000, 2000) = 10 · 6,0000.6 2,0000.4 = 38,664.
Functions of Three Independent Variables Any local extremum of w = f (x, y, z) subject to the constraint g(x, y, z) = 0 will be among the set of points (x0, y0, z0, λ0) which are a solution to the system Whereprovided that all the partial derivatives exist.This is an extension of the two-variable case.
Example Find the extrema of f (x, y, z) = 2x + 4y + 4z,subject to x2 + y2 + z2 = 9. Step 1. Formulate: Maximize w = f (x, y) = 25 – x2 – y2 subject to g(x, y) = 2x + y – 10= 0 Step 2. Form the function F:
Example Find the extrema of f (x, y, z) = 2x + 4y + 4z,subject to x2 + y2 + z2 = 9. Step 1. Formulate: Maximize w = f (x, y) = 2x + 4y + 4z subject to g(x, y) = x2 + y2 – z2= 9 Step 2. Form the function F: Step 3. Find the critical points for F:
Example Find the extrema of f (x, y, z) = 2x + 4y + 4z,subject to x2 + y2 + z2 = 9. Step 1. Formulate: Maximize w = f (x, y) = 2x + 4y + 4z subject to g(x, y) = x2 + y2 – z2= 9 Step 2. Form the function F: Step 3. Find the critical points for F: Solving yields two critical points: (– 1, – 2, – 2, 1) and ( 1, 2, 2, –1)
Example(continued) Step 4. Conclusion. Since there are two candidates, we need to evaluate the function values: f (1, 2, 2) = 2 + 8 + 8 = 18 f (–1, –2, –2) = – 2 – 8 – 8 = –18 We conclude that the maximum of f occurs at (1, 2, 2), and the minimum of occurs at (–1, –2, –2).
Summary • We learned how to use Lagrange multipliers in the equationto find local extrema for two independent variable problems. • We learned how to use Lagrange multipliers in the equationto find local extrema for three independent variable problems.
