� � 21. Maxima and minima: II To see how to maximise and minimise a function on the boundary, let’s conside a concrete example. Let K = { (x,y) | x 2 + y 2 ≤ 2 }. Then K is compact. Let f : K −→ R, be the function f(x,y)= xy. Then f is continuous and so f achieves its maximum and minimum. I. Let’s first consider the interior points. Then �f(x,y)=(y,x), so that (0, 0) is the only critical point. The Hessian of f is 01 Hf(x,y)= .10 d1 = 0 and d2 = −1=0 so that (0�, 0) is a saddle point. It follows that the maxima and minima of f are on the boundary, that is, the set of points C = { (x,y) | x 2 + y 2 =2 }. II. Let g : R2 −→ R be the function g(x,y)= x2 + y2 . Then the circle C is a level curve of g. The original problem asks to maximise and minimise f(x,y)= xy subject to g(x,y)= x 2 + y 2 =2. One way to proceed is to use the second equation to eliminate a variabble The method of Lagrange multipliers does exactly the opposite. Instead of eliminating a variable we add one more variable, traditionalll called λ. So now let’s maximise and minimise h(x,y,λ)= f(x,y) − λ(g(x,y) − 2) = xy − λ(x 2 + y 2 − 2). We find the critical points of h(x,y,λ): y =2λx x =2λy 2= x 2 + y 2 . First note that if x = 0 then y = 0 and x2 + y2 = 0 = 2, impossible. �So x = 0. Similarly one can check that y = 0 and λ = 0. Divide the �1 ��first equation by the second: yx = , xy so that y2 = x2 . As x2 + y2 = 2 it follows that x2 = y2 = 1. So x = ±1 and y = ±1. This gives four potential points (1, 1), (−1, 1), (1, −1), (−1, −1). Then the maximum value of f is 1, and this occurs at the first and the last point. The minimum value of f is −1, and this occurs at the second and the third point. One can also try to parametrise the boundary: �r(t)= √2(cos t, sin t). So we maximise the composition h: [0, 2π] −→ R, where h(t) = 2cos t sin t. As I = [0, 2π] is compact, h has a maximum and minimum on I. When h�(t)=0, we get cos 2 t − sin2 t =0. Note that the LHS is cos 2t, so we want cos2t =0. It follows that 2t = π/2+2mπ, so that t = π/4, 3π/4, 5π/4, and 7π/4. These give the four points we had before. What is the closest point to the origin on the surface F = { (x,y,z) ∈ R3 | x ≥ 0,y ≥ 0,z ≥ 0,xyz = p }? So we want to minimise the distance to the origin on F . The first trick is to minimise the square of the distance. In other words, we are trying to minimise f(x,y,z)= x2 + y2 + z2 on the surface F = { (x,y,z) ∈ R3 | x ≥ 0,y ≥ 0,z ≥ 0,xyz = p }. In words, given three numbers x ≥, y ≥ 0 and z ≥ 0 whose product is p> 0, what is the minimum value of x2 + y2 + z2? Now F is closed but it is not bounded, so it is not even clear that the minimum exists. Let’s use the method of Lagrange multipliers. Let h: R4 −→ R, be the function h(x,y,z,λ)= x 2 + y 2 + z 2 − λ(xyz − p). 2 We look for the critical points of h: 2x = λyz 2y = λxz 2z = λxy p = xyz. Once again, it is not possible for any of the variables to be zero. Taking the product of the first three equations, we get 8(xyz)= λ3(x 2 y 2 z 2). So, dividing by xyz and using the last equation, we get 8= λ3 p, that is 2 λ = . p1/3 Taking the product of the first two equations, and dividing by xy, we get 4= λ2 z 2 , so that 1/3 z = p. So h(x,y,z,λ) has a critical point at 2 (x,y,z,λ)=(p 1/3 ,p 1/3 ,p 1/3 , 1/3 ). pWe check that the point (x,y,z)=(p 1/3 ,p 1/3 ,p 1/3), is a minimum of x2 + y2 + z2 subject to the constraint xyz = p. At this point the sum of the squares is 3p 2/3 . 1/3Suppose that x ≥√3p. Then the sum of the squares is at least 2/31/31/33p. Similarly if y ≥√3por z ≥√3p. On the other hand, the set K = { (x,y,z) ∈ R3 | x ∈ [0, √3p 1/3],y ∈ [0, √3p 1/3],z ∈ [0, √3p 1/3],xyz = p }, is closed and bounded, so that f achieves it minimum on this set, which we have already decided is at (x,y,z)=(p 1/3 ,p 1/3 ,p 1/3),since f is larger on the boundary. Putting all of this together, the point(x,y,z)=(p 1/3 ,p 1/3 ,p 1/3),3 � is a point where the sum of the squares is a minimum. Here is another such problem. Find the closest point to the origin which also belongs to the cone 22 2 x + y = z, and to the plane x + y + z =3. As before, we minimise f(x,y,z)= x2 + y2 + z2 subject to g1(x,y,z)= x2 + y2 − z2 = 0 and g2(x,y,z)= x + y + z = 3. Introduce a new function, with two new variables λ1 and λ2, h: R5 −→ R, given by h(x,y,z,λ1,λ2)= f(x,y,z) − λ1g1(x,y,z) − λ2g2(x,y,z) = x 2 + y 2 + z 2 − λ1(x 2 + y 2 − z 2) − λ2(x + y + z − 3). We find the critical points of h: 2x =2λ1x + λ2 2y =2λ1y + λ2 2z = −2λ1z + λ2 2 22 z = x + y 3= x + y + z. Suppose we substract the first equation from the second: y − x = λ1(y − x). So either x = y or λ1 = 1. Suppose x = y. Then λ1 = 1 and λ2 = 0. In this case z = −z, so that z = 0. But then x2 +y2 = 0 and so x = y = 0, which is not possible. It follows that x = y, in which case z = ±√2x and (2 ±√2)x =3. So 3 3(2 �√2) x == . 22 ±√2 This gives us two critical points: 3(2 −√2) 3(2 −√2) 3√2(2 −√2)P1 =(,, )22 2 3(2 + √2) 3(2+ √2) 3√2(2 −√2)P2 =(, ). 22 , − 2 Of the two, clearly the first is closest to the origin. 4 To finish, we had better show that this point is the closest to the origin on the whole locus F = { (x,y,z) ∈ R3 | x 2 + y 2 = z 2 ,x + y + z =3 }. Let K = { (x,y,z) ∈ F | x 2 + y 2 + z 2 ≤ 25 }. Then K is closed and bounded, whence compact. So f achieves its minimum somewhere on K, and so it must achieve its minimum at P = P1. Clearly outside f is at least 25 on F \ K, and so f is a minimum at P1 on the whole of F . 5 MIT OpenCourseWarehttp://ocw.mit.edu 18.022 Calculus of Several Variables Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.