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 (4∕3,1). 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