The correct answer is: All integer pairs (x,y) with x = number of $10 tickets and y = number of $20 tickets satisfying $y\in\{0,\dots,8\}$ and $20-y\le x\le 28-2y$. There are 45 possible outcomes.
Explanation
Let $x$ = number of \$10 tickets and $y$ = number of \$20 tickets.
Constraints:
- At least 20 tickets: $x+y\ge20$.
- Total cost at most \$280: $10x+20y\le280$, which simplifies to $x+2y\le28$.
- $x,y\ge0$ and must be integers.
Steps:
- Write the inequalities:
- $x+y\ge20$
- $x+2y\le28$
- Combine to bound $y$:
- Subtracting gives $y\le8$, so $y\le8$ and $y\ge0$, hence $y\in\{0,1,\dots,8\}$.
- For each integer $y$ in that range, $x$ must satisfy
- $20-y\le x\le 28-2y$ (and $x\ge0$, but the left bound already enforces positivity for these $y$).
- List (or compute) the integer solutions by plugging $y=0,\dots,8$:
- $y=0:\ x=20,\dots,28$ (9 values)
- $y=1:\ x=19,\dots,26$ (8 values)
- $y=2:\ x=18,\dots,24$ (7 values)
- $y=3:\ x=17,\dots,22$ (6 values)
- $y=4:\ x=16,\dots,20$ (5 values)
- $y=5:\ x=15,\dots,18$ (4 values)
- $y=6:\ x=14,\dots,16$ (3 values)
- $y=7:\ x=13,\dots,14$ (2 values)
- $y=8:\ x=12$ (1 value)
Total integer outcomes: $9+8+\dots+1=45$.
Geometric description (graph): the feasible region is the triangular region bounded by the lines $x+y=20$, $x+2y=28$, and the $x$-axis. Its