Lagrange Multipliers: Finding Max/Min with Constraints

1. 14.8 • Lagrange Multipliers

2. Lagrange Multipliers • Lagrange’s method is used to find the max or min a general function f(x, y, z) that is subject to a constraint g(x, y, z) = k. • This means that the gradient vectors f (x0, y0, z0) and g(x0, y0, z0) must be parallel. • Therefore, if g(x0, y0, z0) 0, there is a number  such that: • The number  in is called a Lagrange multiplier.

3. Lagrange Multipliers • The procedure is as follows:

4. Lagrange Multipliers • Rewriting the vector equation f =  g in terms of components gives: • fx = gxfy = gyfz = gzg(x, y, z) = k • This is a system of four equations in the four unknowns: x, y, z, and . • Let’s work an example…

5. Example 1 • A rectangular box without a lid is to be made from 12 m2 (surface area) of cardboard. Find the maximum volume of such a box. • Solution: • Let x, y, and z be the length, width, and height, respectively, of the box in meters. • We want to maximize the volume: V = xyz • subject to the constraint: g(x, y, z) = 2xz + 2yz + xy = 12 • So using the method of Lagrange multipliers, we look for values of • x, y, z, and  such that V = g and g(x, y, z) = 12.

6. Example 1 – Solution • cont’d • This gives the equations • Vx= gx • Vy= gy • Vz= gz • 2xz + 2yz + xy = 12 • which become: • yz = (2z + y) (1) • xz = (2z + x) (2) • xy = (2x + 2y) (3) • 2xz + 2yz + xy = 12 (4)

7. Example 1 – Solution • cont’d • There are no general rules for solving systems of equations. Sometimes some ingenuity is required. • In the present example you might notice that if we multiply (1) by x,(2) by y, and (3) by z, then the left sides of these equations will be identical. • Doing this, we get: • xyz = (2xz + xy) • xyz = (2yz + xy) • xyz = (2xz + 2yz) • This means: 2xz + xy = 2yz + xy • which gives xz = yz and so x = y.

8. Example 1 – Solution • cont’d • Also: 2yz + xy = 2xz + 2yz • so: 2xz = xy and so y = 2z. • If we now put x = y = 2z in equation (4): 2xz + 2yz + xy = 12 • we get 4z2 + 4z2 + 4z2 = 12 • Solve for z: z = 1 and so x = 2 and y = 2. • Then V = 2  2  1 = 4, so the maximum volume of the box is: 4 m3.

9. Two Constraints

10. Two Constraints • we want to find the maximum and minimum values of a function • (x, y, z) subject to two constraints (phew!): • g(x, y, z) = k and h(x, y, z) = c. • So there are numbers  and  (called Lagrange multipliers) such that • In this case Lagrange’s method is to look for extreme values by solving five equations in the five unknowns x, y, z, , and .

11. Two Constraints • in terms of components and using the constraint equations: • fx = gx + hx • fy = gy + hy • fz = gz + hz • g(x, y, z) = k • h(x, y, z) = c

12. Example 2 • Find the maximum value of the function f (x, y, z) = x + 2y + 3z • on the curve of intersection of the plane x – y + z = 1 and • the cylinder x2 + y2 = 1. • Solution: • We maximize the function f (x, y, z) = x + 2y + 3z subject to the constraints g(x, y, z) = x – y + z = 1 and h(x, y, z) = x2 + y2 = 1. • The Lagrange condition is f = g + h, so the equations to solve are: • 1 =  + 2x • 2 = – + 2y • 3 =  • x – y + z = 1 • x2 + y2 = 1

13. Example 2 – Solution • cont’d • The Lagrange condition is f = g + h, so we solve the equations • 1 =  + 2x (1) • 2 = – + 2y (2) • 3 =  (3) • Therefore: x – y + z = 1 • x2 + y2 = 1 • Putting  = 3 from equation (3) in equation (1) we get 2x= –2, so: x = –1/. Similarly, from (2) we get: y = 5/(2).

14. Example 2– Solution • cont’d • Substitution gives: • so 2 =29/4 so = . • Then x = y = and z = 1 – x + y = 1 • The corresponding values of f are: • Therefore the maximum value of f on the given curve is: