Mathematics 205 HWK 18 Solutions Section 15.3 p725

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1 Mathematics 205 HWK 18 Solutions Section 15.3 p725 Problem 3, 15.3, p725. Use Lagrange multipliers to find the maximum and minimum values, if they exist, of f(x, y) = 3x 2y subject to the constraint x 2 + 2y 2 = 44. Solution. Let g(x, y) = x 2 + 2y 2. The given constraint limits (x, y) to the ellipse g(x, y) = 44, which is a closed, bounded set. Meover, the given function f is continuous everywhere. Therefe we know that f has both a global maximum and a global minimum when (x, y) is held subject to the given constraint. The geometry of the situation is also pretty easy to analyze. We know (x, y) is constrained to lie on the ellipse x 2 + 2y 2 = 44. The contours f f(x, y) = 3x 2y will be straight lines. Two of the contours will be tangent to the ellipse. From the geometry we know that one of these points will give us the desired maximum, the other the desired minimum. Notice that we have no endpoints to wry about. N does g = 0 anywhere on the constraint set (since g = 0 would require x = y = 0, which is impossible if x 2 + 2y 2 = 44). From the sketch, it appears that the maximum occurs somewhere in the fourth quadrant, and the minimum occurs somewhere in the second quadrant. We know each occurs at a point where f λ g = 0. In fact, we know that the maximum and minimum must occur at points where f λ g = 0 and g = 44. From the condition f x λg x = 0 we get the equation From f y λg y = 0 we get 3 2λx = λy = 0. Page 1 of 8 A. Sontag November 11, 2003

2 We also have the constraint equation x 2 + 2y 2 = 44. Note that λ cannot be zero, lest the first equation give us 3 = 0, which is impossible. Solve the first equation f x in terms of λ, solve the second equation f y in terms of λ, and put the results into the third equation. This gives us x = 3 2λ Rewriting a little, we have This gives us so y = 2 4λ = 1 2λ ( 3 2λ )2 + 2( 1 2λ )2 = λ λ 2 = λ 2 = λ 2 = 4. λ 2 = 1 16, λ = ± 1 4. With λ = 1 4, we find x = 6, y = 2, and f(6, 2) = = 22. With λ = 1 4, we find x = 6, y = 2, and f( 6,2) = 18 4 = 22. Subject to the given constraint, therefe, the maximum value f f is 22, occuring at (x, y) = (6, 2), in the fourth quadrant, and the minimum value f f is 22, occuring at the point ( 6, 2), in the second quadrant. Page 2 of 8 A. Sontag November 11, 2003

3 Problem 7, 15.3, p725. Use Lagrange multipliers to find the maximum and minimum values, if they exist, of f(x, y) = x 2 xy + y 2 subject to the constraint x 2 y 2 = 1. Solution. Write g(x, y) = x 2 y 2, so the constraint graph consists of all points f which g(x, y) = 1. This graph is a hyperbola (two branches), which I ll call H. Does f have a maximum value and a minimum value f (x, y) restricted to the hyperbola H? Perhaps not. Although f is continuous everywhere, the constraint hyperbola H is not bounded. So there may not be a maximum value a minimum value. In fact, it s fairly easy to see that as (x, y) moves outward along any of the four tips of the hyperbola, f(x, y) eventually gets very large, and in fact +. F instance, f the tip of the hyperbola that extends up into the outer reaches of the first quadrant, very far out on the hyperbola we will have x > y, x y, x 2 xy+y 2 y 2. So as y + and (x, y) follows that tip of the hyperbola, we will have x 2 xy + y 2 y 2 + as well. Thus f(x, y) takes on arbitrarily large positive values at points of H, so there is no maximum value on the hyperbola H. Alternatively, we could complete the square to write (x 2 xy + y 2 ) = (x y 2 )2 + 3y Letting y +, with (x, y) on the hyperbola, we ll have f(x, y) +, confirming that there is no maximum value f f on H. Page 3 of 8 A. Sontag November 11, 2003

4 There could be a minimum value, however. In fact, there will be. To confirm this, notice that f(1, 0) = 1, so f is comparatively small when x and y are comparatively small. If we just consider the ption of H that lies inside on some huge circle, say the circle x 2 + y 2 = 10 10, then f will have both a maximum and a minimum value on this ption (call it H, say) of the hyperbola. It s pretty clear that the minimum f f on H doesn t lie at any of the endpoints of H (because f is way bigger than 1 out there), so this minimum must be at some point on H that s not an endpoint. Wherever this minimum occurs will also be where the minimum f f on H occurs. Now that we re convinced there is a minimum value, we can use the Lagrange multiplier method to locate it. Setting up the equations f x λg x = 0, f y λg y = 0, g(x, y) = 1, we have 2x y 2λx = 0 x + 2y 2λy = 0 x 2 y 2 = 1. This system needs a little ingenuity (and probably a calculat) to solve. Here s one way to eliminate λ. Multiply the first equation by y (not zero, lest the second equation give us x = 0 which then makes the first equation say 0 = 1, impossible). We will also multiply the second equation by x (not zero, lest the first equation give us y 2 = 1, impossible). Our new system is 2xy y 2 2λxy = 0 2xy x 2 + 2λxy = 0 x 2 y 2 = 1. Add the first two equations. This gives us x 2 + 4xy y 2 = 0, which is equivalent to x 2 4xy + y 2 = 0. Solve this equation f y in terms of x. If you wk it out with the quadratic fmula, you should get y = x(2 ± 3). Now substitute y = x(2 ± 3) into x 2 y 2 = 1 and solve f x. With y = x(2 + 3), we have x 2 x 2 (2 + 3) 2 = 1 [1 (2 + 3) 2 ]x 2 = 1. Since the coefficient 1 (2 + 3) 2 is negative, this is impossible. With y = x(2 3), we have x 2 x 2 (2 3) 2 = 1 Page 4 of 8 A. Sontag November 11, 2003

5 x 2 = 1 1 (2 3) 2 1 x = ± 1 (2. 3) 2 We could solve f y exactly, but this is probably a good time to fego the remaining algebra and dig out a calculat. F decimal equivalents, the calculat gives x ± which gives say y ± x ±1.038, y Notice that the two + signs go together, as do the two signs. Thus, so far we know that the possible locations f the minimum value f f(x, y) = x 2 xy + y 2 on the hyperbola x 2 y 2 = 1 are (approximately) the points (1.038, 0.278) and ( 1.038, 0.278). Evaluate f(x, y) at these points: f(1.038, 0.278) = (1.038) 2 (1.0 38)(0.278) + (0.278) 2 = f( 1.038, 0.278) = Thus the minimum value f f on H is 0.866, and it occurs at both of the points (1.038,0.278) and ( 1.038, 0.278). As a partial check, compare with a few other values: (±1) = 1 and f(±1,0) = 1 > 0.866; (± 2) 2 (±1) 2 = 1 and f(± 2, ±1) = = 3 2 > (This proves nothing, but is reassuring after all that algebra!) Finally, I thought you might like to see a computer plot. On the next page is a graph showing three contours of f, namely f = 0.4, f = 0.866, and f = 1.5, all of which are ellipses. The graph also shows the hyperbola g = 1, which is tangent to the contour f = (the middle one of the ones shown) at the points already identified. Page 5 of 8 A. Sontag November 11, 2003

6 Problem 16, 15.3, p725. The figure shown in the text (Fig p726) shows contours of f and a constraint curve g(x, y) = c. Does f have a maximum value subject to the constraint g(x, y) = c f x 0, y 0? If so, approximately where is it and what is its value? Does f have a minimum value subject to the constraint? If so, approximtely where and what? Solution. From the sketch in the text, it appears that the constraint set g(x, y) = c, x 0, y 0, is tangent to the contour f(x, y) = 400 at approximately (6,6). Meover, the sketch suggests that all the other points meeting the constraints give lower levels f f than 400, so the maximum would be 400, occurring at approximately (6, 6). To find lower values f f on the constraint set, we can move out toward the endpoints. It s not really possible to say whether f has a minimum, but if it does it looks like it would be something less than 100 and occuring at about (10.5,0), an endpoint, and perhaps also at about (0,13.5), the other endpoint. Problem 17, 15.3, p725. (a) Draw contours of f(x, y) = 2x+y f z = 7, 5, 3, 1, 1,3,5, 7. (b) On the same axes, graph the constraint x 2 + y 2 = 5. (c) Use the graph to approximate the points at which f has a maximum a minimum value subject to the constraint x 2 + y 2 = 5. (d) Use Lagrange multipliers to find the maximum and minimum values of f(x, y) = 2x+y subject to x 2 + y 2 = 5. Solution. (a) and (b). The contours f f are straight lines, each with slope 2 and y-intercept equal to the specified z-level. (F instance, the contour z = 7 is the line 2x + y = 7, which is y = 2x 7, so has slope 2 and y-intercept 7.) The constraint curve is a circle, centered at the igin, with radius 5. Here s a sketch. (I haven t labeled the contours, but you can tell what the level is by looking at the y-intercept. F instance the one farthest to the right is f the level z = 7. Page 6 of 8 A. Sontag November 11, 2003

7 (c) It appears that the maximum value is f = 5, occurring at the point (2,1). It appears that the minimum value is f = 5, occurring at ( 2, 1). (d) The geometry makes clear that the required maximum and minimum exist. (Or we could invoke our existence theem. The constraint curve is a closed, bounded set, and the objective function f is continuous on this set. So the required maximum and minimum exist.) The geometry also makes clear that we don t have any endpoints to wry about. Meover, g cannot be 0 on the constraint set (since g = 0 only at the igin). So the maximum and minimum must occur where f λg = 0. The equations to solve are 2 2λx = 0 1 2λy = 0 x 2 + y 2 = 5. Realize that λ cannot be zero (lest the first equation say that 2 = 0, impossible). Solve the first equation f x in terms of λ and the second equation f y in terms of λ: x = 1 λ Page 7 of 8 A. Sontag November 11, 2003

8 Put these results into x 2 + y 2 = 5 to get y = 1 2λ. 1 λ λ = λ 2 = 5 λ 2 = 1 4 which gives us λ = ± 1 2. With λ = 1 2, we find x = 2 and y = 1; we have f(2, 1) = 5. With λ = 1 2, we find x = 2, y = 1; we have f( 2, 1) = 5. As expected, the maximum value is 5, occurring at the point (2,1), and the minimum value is 5, occurring at the point ( 2, 1). Page 8 of 8 A. Sontag November 11, 2003