1. Solve for a
a(a-1)x = a^{2}
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 = a, or (x-1)a= x in standard form.
So a= x/(x-1) when x =/ 1 and there are no nozero solutions when x =1.
Remark. After solving this quadratic equation, you should appriciate how easy liner equations are.
2. xz -> max, |x-1| + |z+ 1|
<= 5.
Solution. The feasible region can be given by 4 linear constraints. It is is diamand with vertices
(x, z) = (6, -1), (1,4), (-4, -1), (1,-6). Answer: max= 6.25 at x = z = 2.5 or x = z = -2.5.
3. Write LP in standard and canonical forms:
f= 2x - y -> max, -1 <= x <=
9, y <= 0.
Solution. Let u = x + 1 >= 0, w = - y >= 0., s = 10 -u = 9 - x >= 0.
Canonical form: -f = -2u - w +2 -> min, u <= 10; u, w > = 0.
Standard form: -f = -2u - w + 2 -> min, u + s = 10; u, w, s > = 0.
4. Solve for x,y:
3 | x | 2 | y | |
x | -1 | y | -2 | = x |
-1 | 0 | x | 0 | = -3 |
x | u | 4 | 3 | |
4 | 4 | 2 | x | = y |
-1 | 2* | 0 | 0 | = v |
x | v |
4 | 3 | |
6 | 2 |
2 | x | = y |
1/2 | 1/2 | 0 | 0 | = u |