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 t**he 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 |

If we start to change the first limit 200 (instead of the second limit 200), then the slope would be 5/4.

Graphical solution.

When 0 <= t <= 100, the first constraint is redundund, the feasible region is a triangle, max = 3t at x=t, y = 0.

When 100 <=t <= 300, the optimal solution is the intersection of the lines corresponding to the first two constraints:

2x+2y = 200, x+3y = t, hence x= 150-t/2, y=t/2 -50, max = 250+t/2.

When t >=300, the second constraint is redundand, the feasible region is a triangle, max = 400 at x=0, y = 100.