5.2 Solutions to exercises on local optima

1.  f '(x) = −3x² + 3 and  f ''(x) = −6x, so  f '(1) = 0 and  f ''(1) = −6 < 0. Thus x = 1 is a local maximizer.

2. 1.  f '₁(x, y) = −2x + y + 2 and  f '₂(x, y) = x − 2y + 1. So the first-order conditions have a unique solution, (x, y) = (5/3, 4/3). We have  f "₁₁(x, y) = −2,  f "₂₂(x, y) = −2, and  f "₁₂(x, y) = 1, so  f "₁₁(5/3, 4/3) f "₂₂(5/3, 4/3) − ( f "₁₂(5/3, 4/3))² = 3 > 0. Hence the Hessian of  f  is negative definite at (5/3, 4/3), so that (x, y) = (5/3, 4/3) is a local maximizer.

   2.  f '₁(x, y) = 2e^2x − 2,  f '₂(x, y) = 4y. So the first-order conditions have a unique solution, (x, y) = (0, 0). We have  f "₁₁(x, y) = 4e^2x,  f "₂₂(x, y) = 4, and  f "₁₂(x, y) = 0, so  f "₁₁(0, 0) f "₂₂(0, 0) − ( f "₁₂(0, 0))² = 16e^2x > 0. Hence the Hessian of  f  is negative definite at (0, 0), so that (x, y) = (0, 0) is a local minimizer.

3. 1. We have  f '(x) = 2x, so the function has a single stationary point, x = 0. We have also  f "(x) = 2 > 0, so x = 0 is a local minimizer. We have also  f (−1) = 4 and  f (1) = 4, so that x = 0 is the only global minimizer (value 3), and x = −1 and x = 1 are both global maximizers (value 4).
   2. We have  f '(x) = 3x² − 3, so the function has two stationary points, x = 1 and x = −1. We have  f "(x) = 6x, so that  f "(1) > 0 and  f "(−1) < 0. Thus x = 1 is a local minimizer (minimum value 3) and x = −1 is a local maximizer (maximum value 7). We have also  f (−3) = −13 and  f (3) = 23, so x = −3 is the only global minimizer (value −13) and x = 3 is the only global maximizer (value 23).
   3. We have  f '(x) = 1 − 1/x², so the function has a single stationary point in the interval [1/2, 2], namely x = 1. We have  f "(x) = 2/x³, so x = 1 is a local minimizer (value 2). We have also  f (1/2) = 5/2 and  f (2) = 5/2, so x = 1 is the only global minimizer (value 2) and x = 1/2 and x = 2 are both global maximizers (value 5/2).
   4. We have  f '(x) = 6(x − 2)⁵, so the function has a single stationary point, x = 2. We have  f "(x) = 30(x − 2)⁴, so that  f "(2) = 0. Thus this information is not enough to tell us whether x = 2 is a local maximizer or minimizer, or neither. We have also  f (0) = 64 and  f (4) = 64, so x = 2 is the only global minimizer (value 0), and x = 0 and x = 4 are both global maximizers (value 64). We conclude that x = 2 is the only local minimizer, and x = 0 and x = 4 are both local maximizers.

4. 1. The first-order conditions are

      x2 + 2y − 6	=	0
      2x − 4y	=	0.

      From the second equation we have x = 2y, so from the first equation we have

      4y² + 2y − 6 = 0,

      or

      2y² + y − 3 = 0,

      or

      (2y + 3)(y − 1) = 0.

      Thus y = 1 or y = −(3/2). Hence the equations have two solutions: (2, 1) and (−3, −3/2).

      The Hessian of the function is

      2x   2   2   −4

      .

      For x = 2 this matrix is indefinite;  f "_x(2, 1) > 0 and  f "_y(2, 1) < 0, so (2, 1) is a saddle point. For x = −3 the matrix is negative definite, so that (−3, −3/2) is a local maximizer.

      The point (−3, −3/2) is not a global maximizer, because for y = 0 and x arbitrarily large, the value of the function is arbitrarily large.

   2. The first-order conditions are

      3y − 3x2	= 0
      3x − 3y2	= 0.

      These equations have two solutions, (0, 0) and (1, 1).

      The Hessian of the function is

      −6x   3   3   −6y

      At (1,1) this Hessian is negative definite, while at (0,0) it is indefinite.

      Thus (1,1) is a local maximizer. It is not a global maximizer, because  f (1, 1) = 1, while  f (−1, −1) = 5 (for example).

5. 1. In the figure below, the function is positive in the red region (i.e. when x > y² and x > 2y², or x < y² and y < 2y²) and negative in the green region.

      

   2. We have g_a(y) =  f (ay, y) = (y − a²y²)(y − 2a²y²) = y² − 3a²y³ + 2a⁴y⁴. The derivative is 2y − 9a²y² + 8a⁴y³, which is zero when y = 0. The second derivative is 2 − 18a²y + 24a⁴y², which is positive at y = 0, so for any value of a the function has local minimum at 0.
   3. No: look at the diagram for part (a). Every disk centered at (0, 0), not matter how small, contains points at which the function is negative (points between the two parabolas).

6. 1. First find the stationary points: these are the solutions of  f '(x) = 0, or 9x² − 10x + 1 = 0, or (9x − 1)(x − 1) = 0. Thus the stationary points are x = 1/9 and x = 1. The values of the function at these points are  f (1/9) = 3(1/9)³ − 5(1/9)² + 1/9 = (1/243)[1 − 15 + 27] = 13/243 and  f (1) = −1.

      Now find the values of the function at the endpoints of the interval:  f (0) = 0 and  f (1) = −1.

      Thus the (global) maximizer of the function on the interval [0, 1] is x = 1/9 and the (global) minimizer is x = 1. (There are no other local maximizers or minimizers.)

   2. As in the previous part, the stationary points are x = 1/9 and x = 1, with  f (1/9) = 13/243 and  f (1) = −1.

      The values of the function at the endpoints of the interval are  f (0) = 0 and  f (2) = 6.

      Thus the (global) maximizer of the function on the interval [0, 2] is x = 2 and the (global) minimizer is x = 1.

      The only other stationary point is x = 1/9. We have  f "(x) = 18x − 10, so that  f "(1/9) < 0. Thus x = 1/9 is a local maximizer.

   3. We have  f '(x) = 2(x − 4) and hence  f "(x) = 2 for all values of x. Thus the function  f  is convex. Hence any stationary point is a global minimizer.

      There is a single stationary point, x = 4. Thus the function has a single global minimizer, x = 4.

      We have  f (−5) = 86 and  f (5) = 6, so the global maximizer is x = −5.

7. 1.  f (2, 3) = 0 ≤ (x − 2)⁴ + (y − 3)⁴ for any x and y, because z⁴ ≥ 0 for all z.

   2.  f ₁'(x, y) = 4(x − 2)³,  f ₂'(x, y) = 4(y − 3)³. So the only solution of first-order conditions is (x, y) = (2, 3).

   3.  f ₁₁"(x, y) = 12(x − 2)²,  f ₂₂"(x, y) = 12(y − 3)², and  f ₁₂"(x, y) = 0, so that we have  f ₁₁"(2, 3)· f ₂₂"(2, 3) − ( f ₁₂"(2, 3))² = 0. Thus the Hessian is not positive definite at any solution of the first-order conditions. (Hence we cannot conclude from the second-order conditions that (2, 3) is a minimizer of the function, though we know that it is.)

8. 1. First-order conditions:

      f 1'(x1, x2, x3) = 2x1 − 3x2	=	0
      f 2'(x1, x2, x3) = −3x1 + 6x2 + 4x3	=	0
      f 3'(x1, x2, x3) = 4x2 + 12x3	=	0

      These conditions have a unique solution, (x₁, x₂, x₃) = (0, 0, 0). The Hessian matrix is

      2   −3   0   −3    6   4   0   4   12

      The leading principal minors are 2 > 0, 3 > 0, and (2)(72−16)−(−3)(−36) = 4 > 0, so that (x₁, x₂, x₃) = (0,0,0) is a local minimizer. (The function is convex, so this minimum is in fact a global minimizer.)
   2. First-order conditions:  f _i'(x₁, x₂, x₃) = −2x_i = 0 for i = 1, 2, 3. There is a unique solution, (x₁, x₂, x₃) = (0,0,0). The Hessian matrix is

      −2   0   0   0   −2   0   0   0   −2

      The leading principal minors are −2 < 0, 4 > 0, and (−2)(4) = −8 < 0, so that (x₁, x₂, x₃) = (0,0,0) is a local maximizer.

   3. First-order conditions:

      f 1'(x1, x2, x3) = 2x1 + x3	=	0
      f 2'(x1, x2, x3) = 2x2 + x3 − 1	=	0
      f 3'(x1, x2, x3) = x1 + x2 + 6x3	=	0

      These conditions have a unique solution, (x₁, x₂, x₃) = (1/20, 11/20, −2/20). The Hessian matrix is

      2   0   1   0   2   1   1   1   6

      The leading principal minors are 2 > 0, 4 > 0, and (2)(11) + (1)(−2) = 20 > 0, so that (x₁, x₂, x₃) = (1/20, 11/20, −2/20) is a local minimizer.