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Math 484.2  October 5, 2004. 
Name:_________________________________________
Midterm 1,  5 problems, 15 points each.

1. Solve for a
 a(a-1)x=  a
where x is a given number.
Solution.  a = 0 is a solution for every x.  To find other solutions,  divide the equation by a.
Then we get a linear equation    (a-1)x =  1, or   xa=1+ x in standard form.
So   a= (1+x)/x  when  x =/ 0 and  there are  no nozero solutions when x =0.
Remark. After solving this quadratic equation, you should appriciate how easy  liner equations are.

2.  yz -> min, |y-1| +  |z+ 3|  <=  5.
Solution.  The feasible region ia a diamond which can be defined by 4 linear constraints.
The vertices are : y = 1 ,z = 2; y = 1, z=-8; y = 6, z = -3;  y = -4, z= -3. 
Answer:   min = -20.25 at  y = -z = 4.5.

3. Write LP in standard and canonical forms:

f=  x+ 2y -> max,  1 <= x <= 5, y <= 0.
  
Solution.  Set   u = - y >= 0, v= x -1 >= 0. w = 5 -x >= 0.
Canonical form: -f = -x +2u -> min,  -x <= -1, x <= 5; x,u >= 0
Standard form:  -f = -x +2u -> min,  -x +v = -1, x + w= 5; x,u, v, w >= 0
Smaller canonical form: -f   =  -v+2u -1 -> min,  v <= 4; u,v >=  0.
Smaller standard form: -f   =  -v+2u -1 -> min,  v +w= 4; u,v ,w >=  0.


 4. Solve for x,y:
 
2x2y
x1y1= x
13-x0= 3
Solution. x=1, y= -2/3.

  5. Pivot:  

xyu3
452*x= y
0200= x

Solution.
xyy3
-2-5/21/2-x/2= u
0200= x