Chapter 9.6, Problem 49E

To determine

To calculate the day

Answer to Problem 49E

The attendance of two movies is same on day 5 and day 35.

Explanation of Solution

Given information: the attendances(y) for two movies is,

  $\begin{array}{l}y=-x^2+35x+100\\ y=-5x+275\end{array}$

Where x is number of days since the movies opened.

Formula used:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

Movie 1 is $y=-x^2+35x+100$

Movie 2 is $y=-5x+275$

Substituting value of equation 1 in equation 2,

  $-x^2+35x+100=-5x+275$

Shifting the values,

  $-x^2+35x+100+5x-275=0$

On solving,

  $-x^2+40x-175=0$

Hence,

  $a=-1$ ,

  $b=40$ ,

  $c=-175$ .

Substituting the formula,

  $x=\frac{-40\pm \sqrt{{40}^2-4(-1)(-175)}}{2(-1)}$

So,

  $x=\frac{-40\pm \sqrt{1600-700}}{-2}$

On solving,

  $x=\frac{-40\pm \sqrt{900}}{-2}$

So,

  $x=\frac{-40\pm 30}{-2}$

For +30,

  $x=\frac{-40+30}{-2}$

On solving,

  $x=5$

Value of y, substituting in equation 2,

  $y=-5(5)+275$

On solving,

  $y=250$

For -30,

  $x=\frac{-40-30}{-2}$

On solving,

  $x=35$

Value of y, substituting in equation 2,

  $y=-5(35)+275$

On solving,

  $y=100$

Hence, $x=5,35$ and $y=250,100$

Hence, the attendance of two movies is same on day 5 and day 35.