Math 484.1  February. 9, 2005    Name:_________________________________________
Midterm 1,  5 problems, 10 points each.  Return this page.

1. Solve for x
x(x+1)a =  x2
where a is a given number.
Solution. Clearly, x = 0 is a solution for every a. To find other solutions, we assume that  x is not 0
and divide the equation by x. We obtain a linear equation  (x+1)a =  x. In standard form,
x(a-1)=-a. Answer: If a is not 1, then   x = 0 or a/(1-a); if  a = 1, then  x = 0.

2. Solve for x, y
 x
  y
 3
 -1
 = a
-8
  3
 = b
where a,b are given numbers.
Solution.  We make two pivot steps:
 x
  y
 3
 -1*
 = a
-8
  3
 = b
 x
  a
 3
 -1
 = y
1*
  -3
 = b
 b
  a
 3
 8
 = y
1
  3
 = x


3. Solve:   xy -> min,    |x|  <=  1,  |y| <= 2.
Solution. min = -2 at  x=-1, y = 2  or  x = 1, y = -2.

4-5. Solve linear programs where all xi >= 0:

-2 x2 1
 -x3 Problem 4
 1 2 3  4 = x1
 1 0 -x1  1 = -x4
 0 1 2  0 -> min

Solution.   The first row says
2x2-4x3+1=x1.
The second row says
-x1-x3-- 2= -x4 hence x1+ x3-+ 2= x4.
Plugging  from the first equation to the second eqution, we get
2x2-4x3+1 + x3-+ 2= x4   hence  2x2-3x3 + 3= x4
We rewrite the LP in a standard tableau: 

x2
 x3
 1
2
 -4
 1
 = x1
2
 -3
 3
 = x4
1
 0
 2
 -> min

The tableau is optimal, so min = 2 at   x2 = x3 = 0,   x1 = 1,  x4= 3.

 2 x2 3 -x1 Problem 5
 1 2 3  4 = -x1
 5 6 0  0 = x4
 0 1 2 -1 -> min

 We combine the columns with constants 2 and 3 on top :

x2
 -x1
 1
2
 4
 11
 = -x1
6
 0
 10
 = x4
1
 -1
 6
 -> min

Next we pivot on 2:

-x1
 -x1
 1
1/2
 -2
 -11/2
 = x2
3
 -12
 -23
 = x4
1/2
 -3
 1/2
 -> min

Now we combine the first two columns and obtain a standard tableau: 

x1
 1
3/2
 -11/2
 = x2
9
 -23
 = x4
5/2
 1/2
 -> min

In other terms,  3x1/2 -11/2 >= 0, 9x1 -23 >=0, 5x1/2 +1/2 -> min,  x1>= 0.
The feasible region is   x1>= 11/3.  So  min = 55/6+1/2 = 29/3 at
x1= 11/3.  The values for the other variables are    x2   = 0, x4 = 10 .
An optimal tableau can be obtained from the standard tableau by pivoting on 3/2.