1. Solve for x,y:
x 
y 
z 

1 
2 
0 
= u 
1 
0 
2 
= v 
Solution — Pivoting

> 

> 

x = v – 2z
y = v/2 + u/2 + z
2. 4x + y^2 –> min
x^2 + y^2 = 1
Solution
solve eq. 2 for y^2:
y^2= 1 – x^2
substitute into eq. 1:
4x + (1 – x^2) –> min
x^2 + 4x + 1 –> min
Objective funcion is now defined only by x
Eq. 2 is the formula for a circle of radius 1, so our feasible range of x is 1
<= x <= 1.
The objective function is minimal when x = 1, with an optimal solution value of 4.
3. Solve for x,y
x 
y 
1 

1 
1 
x 
>= 3 
1 
1 
2 
<= 22 
1 
2 
0 
–> min 
Solution: write
equation from row 1:
x + y – x >= 3
y >= 3
From row 2:
x + y + 2 <= 2
x +y <= 0
y <= x
x >= y
Combining the equations, x >= y >= 3
Objective function is 2x + y, so graphical method can be used or
it should be obvious that minimal values of x and y will result in
minimal objective function, so:
min = 9 when x = y = 3
4. Solve the linear programs given by the following standard tableaux:
4A: 

Tableau is Optimal 
4B: 

Tableaux is
infeasible 
4C: 

Tableau (The LP) is unbounded 
5. Find all logical implications between the following 5 constraints on x,
y:
(a) x_ = y_
(b) 0 > 1
(c) 2y > x
(d) ^{1}/_{3} > ^{1}/_{2}
(e) y >= x
Solution:
(b) is unconditionally true, so (b) is implied by anything
(d) is unconditionally false, so (d) implies everything
a => b
c => b
d => a
d => b
d => c
d => d
d => e
e => b
a => e (added by L.V.)