A radio station is giving away tickets to a play. They plan to give away tickets for seats that cost $10 and $20. They want to give away at least 20 tickets. The total cost of all the tickets they give away can be no more than $280.

All integer solutions (let $x$ = number of $10 tickets$, $y$ = number of $20 tickets$) are:

  • For $y=0,1,\dots,8$, any integer $x$ with $20-y \le x \le 28-2y$.

There are 45 possible combinations. The maximum number of tickets is 28 (all $10$-dollar tickets); the maximum number of $20$-dollar tickets is 8 (with $x=12$).

Explanation

  1. Write the constraints:
  • At least 20 tickets: $x+y \ge 20$.
  • Cost no more than \$280: $10x+20y \le 280$.
  • $x,y$ are nonnegative integers.
  1. Simplify the cost inequality by dividing by 10:
  • $x+2y \le 28$.
  1. Combine inequalities to bound $y$:
  • Subtract $(x+y \ge 20)$ from $(x+2y \le 28)$ to get $y \le 8$.
  • So $y=0,1,\dots,8$.
  1. For each fixed integer $y$ in that range, solve for $x$:
  • From $x+y \ge 20$ we get $x \ge 20-y$.
  • From $x+2y \le 28$ we get $x \le 28-2y$.
  • So $20-y \le x \le 28-2y$ (integers).
  1. Count solutions:
  • For $y=0$ there are $9$ values ($x=20\ldots28$),

for $y=1$ there are $8$ values, …, for $y=8$ there is $1$ value.

  • Total $9+8+\dots+1 = 45$ combinations.

Examples of extremes:

  • Maximum tickets: $x=28,y=0$ gives 28 tickets costing \$280.
  • Maximum $20$-dollar tickets: $y=8,x=12$ gives 20 tickets costing \$280.
  • Minimum cost while meeting at least 20 tickets: $x=20,y=0$ costing \$200.
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