Math 509: Advanced Analysis

Homework 3
Lawrence Tyler Rush
<me@tylerlogic.com>
February 7, 2015
http://coursework.tylerlogic.com/courses/upenn/math509/homework03

The following solutions correspond to Frank Jones’ book, chapter three: http://www.owlnet.rice.edu/~fjones

1 Problem 3-18


2 Problem 3-20


Let f : R2 R be defined as
f (x,y) = 1x3 + 1 y2 + 2xy+ 5x +y
         3    2

This means the partial derivatives are

fx = x2 + 2y + 5 and   fy = y+ 2x+ 1
so that fx is zero at points satisfying
    - 1 2
y = 2 (x + 5)

and fy is zero at all points satisfying

y = - 2x - 1

These constraints yield two critical points, (1,-3) and (3,-7). The Hessian matrix is

(  fxx  fxy )   ( 2x  2 )
   fyx  fyy   =   2   1

Therefore

                 (      )
det(H(1,- 3)) = det 2 2   = - 2
                    2 1

indicating that (1,-3) is a saddle point. Furthermore

                 ( 6  2 )
det(H (3,- 7)) = det 2 1   = 2

indicating that (3,-7) is a local minimum since the diagonals of the above Hessian matrix at (3,-7) are possitive.

3 Problem 3-21


Let f : R2 R be defined as
f(x,y) = x2y2 - 5x2 - 8xy - 5y2

This means the partial derivatives are

        2                        2
fx = 2xy - 10x- 8y   and   fy = 2x y - 8x - 10y
so that fx is zero at points satisfying
        2    -1
x = 4y(y - 5)
(3.1)

Substituting this into the equation 0 = 2x2y - 8x- 10y which is obtained by setting fy = 0, we have the following sequence of equations

2y(4y(y2 - 5)-1)2 - 8(4y(y2 - 5)-1)- 10y = 0
      3  2    -2       2    -1
    32y (y - 5)  - 32y(y - 5)  - 10y  =  0
    16y3(y2 - 5)-2 - 16y(y2 - 5)-1 - 5y = 0
        16y3 - 16y(y2 - 5)- 5y(y2 - 5)2 =  0
       y(16y2 - 16(y2 - 5)- 5(y2 - 5)2) = 0
      (   2     2           2    2)
     y 16y - 16y +(5(16)- 5(y  - 5))  =  0
                 y 5(16)- 5(y2 - 5)2   =  0
                    y (16 - (y2 - 5)2)  =  0
           (     2    )(     2    )
          y 4 + (y  - 5() 4 -)(y(  - 5))  =  0
                    y y2 - 1 9- y2   =  0
Thus fy is zero whenever y = 0,±1,±3. By plugging these values back into equation 3.1 yields critical points at (0,0), (1,-1), (-1,1), (3,3), and (-3,-3). The Hessian matrix is
(         )   (                   )
  fxx  fxy       2y2 - 10 4xy- 8
  fyx  fyy  =    4xy- 8   2x2 - 10

With this, we have the following values of the Hessian matrix at the critical points:

               (           )
det(H (0,0)) = det  - 10 - 8   = 36
                  - 8  - 10

                 (  - 8  - 12 )
det(H(1,- 1)) = det - 12 - 8   = - 80

                 (           )
                    - 8  - 12
det(H(- 1,1)) = det  - 12 - 8   = - 80

                (        )
det(H (3,3)) = det  8   28   = - 720
                  28  8

                  ( 8   28 )
det(H (- 3,- 3)) = det 28 8    = - 720

indicating that (0,0) is a local maximum whilst the other four critical points are saddle points.

4 Problem 3-22


Let f : R2 R be defined as
f(x,y) = xy (12 - 3x - 4y)

This means the partial derivatives are

fx = y(12- 6x - 4y) and   fy = x(12- 3x - 8y)
so that fx is zero whenever y = 0 or when (x,y) satisfies
3x = 6- 2y

and fy is zero whenever x = 0 or when (x,y) satisfies

3x = 12- 8y

Combining these four constraints yields critical points of (0,0), (0,3), (4,0), and (43,1). The Hessian matrix is

(  fxx  fxy )   ( - 6y         12- 6x- 8y )
   fyx  fyy   =   12 - 6x - 8y - 8x

With this, we have the following values of the Hessian matrix at the critical points:

                (        )
                  0   12
det(H (0,0)) = det  12  0    = - 144

                (           )
det(H(0,3)) = det - 12 - 12   = 144
                  - 12 0

                ( 0    - 12 )
det(H(4,0)) = det - 12 - 32   = 144

                  (            )
                    - 6 - 4
det(H(4∕3,1)) = det - 4 - 32∕3   = 48

indicating that (0,0) is a saddle point and the remaining three points are local minimums.

5 Problem 3-23


6 Problem 3-24


7 Problem 3-27


8 Problem 3-28