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Zorluk: OrtaAlgebraic Word Problems and Modeling

A theater sells student tickets for $20\$20 each and adult tickets for $35\$35 each. For a specific performance, the theater sold a total of 150150 tickets and collected total revenue of RR dollars. If at least 4040 student tickets were sold and at most 9090 adult tickets were sold, which of the following values could be the total revenue RR? Select all such values.

  1. A
    $2850\$2{}850
  2. $3450\$3{}450Cevap
  3. $4200\$4{}200Cevap
  4. D
    $4400\$4{}400
  5. E
    $4650\$4{}650

Cevap

The possible values for the total revenue RR are $3450\$3{}450 and $4200\$4{}200.
The linear revenue model is R=5250−15sR = 5250 - 15s. Considering both conditions (s≥40s \ge 40 and a=150−s≤90  ⟹  s≥60a = 150 - s \le 90 \implies s \ge 60), the valid range for student tickets is 60≤s≤15060 \le s \le 150. This restricts the possible revenue RR to multiples of $15\$15 between $3000\$3{}000 and $4350\$4{}350. Both $3450\$3{}450 and $4200\$4{}200 fall within this valid range and correspond to integer ticket quantities (s=120,a=30s = 120, a = 30 and s=70,a=80s = 70, a = 80, respectively).

Adım Adım Çözüm

1
Set up equations for the total number of tickets and revenue.
Let ss be the number of student tickets and aa be the number of adult tickets. Then s+a=150s + a = 150, so a=150−sa = 150 - s. Total revenue R=20s+35a=20s+35(150−s)=5250−15sR = 20s + 35a = 20s + 35(150 - s) = 5250 - 15s.
Expressing revenue in terms of a single variable ss simplifies finding the domain and range.
2
Determine the constraints on the variable ss.
We are given s≥40s \ge 40 and a≤90a \le 90. Substituting a=150−s≤90a = 150 - s \le 90 yields s≥60s \ge 60. Combining constraints gives 60≤s≤15060 \le s \le 150.
The number of adult tickets being at most 9090 forces the number of student tickets to be at least 6060.
3
Calculate the upper and lower bounds for the revenue RR.
Maximum revenue occurs when s=60s = 60: Rmax=5250−15(60)=$4350R_{\text{max}} = 5250 - 15(60) = \$4{}350. Minimum revenue occurs when s=150s = 150: Rmin=5250−15(150)=$3000R_{\text{min}} = 5250 - 15(150) = \$3{}000.
Since R=5250−15sR = 5250 - 15s is a decreasing linear function of ss, the maximum revenue occurs at the minimum valid value of ss and vice versa.
4
Evaluate the given choices against the range and divisibility requirements.
RR must be an integer multiple of 1515 subtracted from 52505250, meaning RR must be between $3000\$3{}000 and $4350\$4{}350 inclusive, and (5250−R)(5250 - R) must be divisible by 1515. $3450\$3{}450 (where s=120s = 120) and $4200\$4{}200 (where s=70s = 70) are both valid. $2850\$2{}850 is below the minimum bound, $4400\$4{}400 does not yield an integer value for ss, and $4650\$4{}650 violates the adult ticket upper bound.
Only options meeting both inequality constraints and integer ticket requirements are valid.

Anahtar Kavram

Linear modeling of word problems under linear system constraints and inequalities
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Algebraic Word Problems and Modeling

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