Solution by simplex method.
We introduce two slack variables
u=100-x-y >= 0 and v= t-x-3y >= 0.
We write the problem in a standard row tableau
x | y | 1 | |
-1 | -1 | 100 | =u |
-1 | -3 | t | =v |
-3 | -4 | 0 | =-P->min |
When t < 0, the v-row is bad, so the problem is infeasible.
Assume now that t >0. The tableau is feasible, so we go to Phase 2.
We chose the y-column as the pivot column.
When t >= 300, then we pivot on -1 and obtain
x | u | 1 | |
-1 | -1 | 100 | =y |
2 | 3 | t-300 | =v |
1 | 4 | -400 | =-P->min |
Assume now that 0 < t < 300. Then according to the simplex method we pivot on -3 and obtain
x | v | 1 | |
-2/3 | 1/3 | 100-t/3 | =u |
-1/3 | -1/3 | t/3 | =y |
-5/3 | 4/3 | -4t/3 | =-P->min |
Now the x-column is the pivot column. We have to compare
(100-t/3)/(-2/3) and (t/3)/(-1/3) to choose a pivot entry.
When 0 < t <=100, we pivot on -1/3 and obtain
y | v | 1 | |
2 | 1 | 100-t | =u |
-3 | -1 | t | =x |
5 | 3 | -3t | =-P->min |
u | v | 1 | |
-3/2 | 1/2 | 150-t/2 | =x |
1/2 | -1 | t/2-50 | =y |
5/2 | 1/2 | -t/2-250 | =-P->min |
Note that the slope is decreasing (the law of diminishing return):
interval | 0<=t<=100 | 100 <= t <= 300 | 300 <= t |
slope | 3 | 1/2 | 0 |
Graphical solution.