Since opening night, attendance at Play A has increased steadily, while attendance at Play B first rose and then fell. Equations modeling the daily attendance y at each play are shown below, where x is the number of days since opening night. On what day(s) was the attendance the same at both plays? What was the attendance?
Play A: y = 16x + 150
Play B: y = –x2 + 60x – 10
A. The attendance was the same on day 40. The attendance was 790 at both plays that day.
B. The attendance was the same on day 4. The attendance was 214 at both plays that day.
C. The attendance was the same on days 4 and 40. The attendance at both plays on those days was 214 and 790 respectively.
D. The attendance was never the same at both plays.
To find the day(s) when attendance was the same at both plays, we need to set the equations for Play A and Play B equal to each other and solve for x:
16x + 150 = –x^2 + 60x – 10
Rearrange and simplify:
x^2 + 44x - 160 = 0
Factor:
(x + 10)(x - 16) = 0
Therefore, the attendance was the same on day 16 (since that is the only positive solution). To find the attendance, we can plug x = 16 into either equation:
Play A: y = 16(16) + 150 = 406
Play B: y = –16^2 + 60(16) – 10 = 406
Therefore, the attendance at both plays on day 16 was 406.
The answer is not A, B, or D, but C.
To find the day(s) when the attendance was the same at both plays, we need to find the values of x when the equations for Play A and Play B are equal.
Setting the two equations equal to each other, we have:
16x + 150 = -x^2 + 60x - 10
Rearranging the equation to standard form, we have:
x^2 + 44x - 160 = 0
Factoring the equation, we have:
(x - 4)(x + 40) = 0
Setting each factor equal to zero, we get:
x - 4 = 0, or x + 40 = 0
Solving each equation, we find:
x = 4, or x = -40
Since we are looking for the number of days since opening night (x cannot be negative), the only valid solution is x = 4.
Therefore, the attendance was the same on day 4. To find the attendance, we substitute x = 4 into either equation:
For Play A: y = 16x + 150
y = 16(4) + 150
y = 64 + 150
y = 214
For Play B: y = -x^2 + 60x - 10
y = -(4)^2 + 60(4) - 10
y = -16 + 240 - 10
y = 214
Therefore, the attendance was the same on day 4, and the attendance at both plays was 214 on that day.
The correct answer is option B. The attendance was the same on day 4, and the attendance was 214 at both plays that day.