Lagrange Multipliers In Example 3 of the previous section, we maximized the volume V(l, w, h) = lwh subject to the constraint that A(l, w, h) = 2hl + 2hw + lw = 12. We solved that problem by substituting the constraint into the function we wished to maximize and then found critical points. In this section, we present a second method for dealing with these problems known as Lagrange’s method for maximizing (or minimizing) a general function f(x, y) subject to a constraint g(x, y) = c. This applies to a more general setting than the method we used in the previous section. (Below, we present the theory for two variables, but the method applies to functions of three or more variables, like Example 3 from the last section.) Suppose we want to find the extreme values of f(x, y) subject to a constant g(x, y) = c. That is, we want to find the extreme values of f(x, y) when the point (x, y) is restricted to the level curve g(x, y) = c. Figure 1 below shows some level curves of f(x, y) (in blue) as well as the constraint g(x, y) = c (in red). 12345670.2 0.4 0.6 0.8 1x0.20.4yx0,y0 Figure 1: Level curves of f(x, y) and the constraint g(x, y) = c To maximize f(x, y) subject to g(x, y) = c is the same as finding the largest c value such that the level curve f(x, y) = k intersects g(x, y) = c. In the above figure, this happens at the point (x0, y0). Notice that at this point, the curves just touch each other. That is, they have a common tangent line. (Otherwise, the value of k could be increased further.) But if the two curves have a common tangent line, this means that their gradient vectors must be parallel. (See the colored arrows above.) This gives us the relationship “f(x, y) = l“g(x, y), for some scalar l. 1Using this relationship, we can describe the method of Lagrange multipliers. Method of Lagrange Multipliers To find the maximum and minimum values of f(x, y) subject to the constraint g(x, y) = c (assuming these extreme values exist) 1. Find all values of x, y and l such that “f(x, y) = l“g(x, y) and g(x, y) = c 2. Evaluate f(x, y) at all of the points found in (1). The largest of these values is the maximum value of f(x, y) and the smallest value is the minimum value of f(x, y). Let us return to Example 3 from the last section, this time solving the problem using Lagrange multipliers. Example 1: A cardboard box without a lid is to have a surface area of 12 m2. Find the dimensions that maximize the volume of the cardboard box. Solution: As in Example 3 of the previous section, we let l, w, h be the length, width, and height of the box, respectively. We wish to maximize V(l, w, h) = lwh subject to the constraint A(l, w, h) = 2lh + 2wh + lw = 12. Using Lagrange multipliers, we want to find l, w, h, and l such that “V(l, w, h) = l“A(l, w, h) and A(l, w, h) = 12. This gives us the equations: Vl = lAl, Vw = lAw, Vh = lAh, 2lh + 2wh + lw = 12 which become (i) wh = l(2h + w) (ii) lh = l(2h + l) (iii) lw = l(2l + 2w) (iv) 2lh + 2wh + lw = 12 2There is no general rule for solving these systems of equations. Sometimes some ingenuity is required. Here, notice that if we multiply (i) by l, (ii) by w, and (iii) by h, then the left sides of these equations will be the same. That is, we have: (v) lwh = l(2lh + lw) (vi) lwh = l(2wh + lw) (vii) lwh = l(2lh + 2wh) Notice that l ∫ 0, since if it did, then wh = 0, lh = 0, and lw = 0 from (i), (ii), and (iii) which would contradict (iv). Thus, we can set (v) and (vi) equal and divide through by l to get 2lh + lw = 2wh + lw. Subtracting lw from both sides and then dividing by 2, we see that lh = wh. But because h ∫ 0 (since this would imply that V = 0), we have that l = w. Setting (vi) and (vii) equal and dividing through by l we get 2wh + lw = 2lh + 2wh. Subtracting 2wh from both sides, we have lw = 2lh. Since l ∫ 0 (since this would imply that V = 0), we have that w = 2h. Thus, we have that l = w = 2h. Substituting those values into the constraint equation, 2lh + 2wh + lw = 12, we have 2(2h)h + 2(2h)h + (2h)(2h) = 12h2 = 12. Since h must be positive, we see that h = 1. This implies that l = 2 and w = 2, which are the same values we found previously. Example 2: Find the smallest value of x2 + y2 subject to the constraint y + 3x = 3. Solution: Notice that “f = 2xi + 2yj and “g = 3i + j. Using Lagrange multipliers, we have “f = l“g, where l is a scalar. This gives us the three equations (i) 2x = 3l, (ii) 2y = l, and (iii) y + 3x = 3. Solve for x and y in (i) and (ii), respectively, we have x = (3/2)l and y = l/2. Plugging these into (iii), we have 333522      3. So l = 3/5. Plugging in l = 3/5 into our equations above, we see that x = 9/10 and y = 3/10. Let us examine a contour diagram (Figure 2) to see why this point (0.9, 0.3) is a global minimum and why the function does not have a global maximum. 30.30.60.90.2 0.4 0.6 0.8 1x0.20.40.60.81y0.9,0.3 Figure 2: Level curves of x2 + y2 and the constraint y + 3x = 3 Notice that as the line y + 3x = 3 moves away from the point (0.9, 0.3), the contour lines it will encounter have larger and larger z-values. Hence, the point (0.9, 0.3) is a global minimum and the function does not have any global maximum value. Examples 1 and 2 could have also been solved using the techniques from the last section. That is, we could have solved for one of the variables and substituted it into the function we wished to maximize (minimize). But there are some instances where we cannot solve for one of the variables in the constraint. Our previous methods would fail but the method of Lagrange multipliers will work just fine. Consider the following example. Example 3: Find the maximum and minimum values of the function f(x, y) = x2 + 2y2 that lie on the circle x2 + y2 = 1. Solution: Figure 3: Sketch of x2 + 2y2 and the constraint x2 + y2 = 1 4Figure 3 above shows a sketch of the surface of f(x, …