Question

day 4: min & max word problems 1
chapter 3: quadratic relations

1. the thunderbirds professional indoor soccer team has 900 season ticket holders. the management of the team wants to increase the current price of $400. a survey indicated that for every $20 increase in price, the team will lose 15 season ticket holders. what price would maximize revenue from season ticket holders? what is the maximum revenue the team could receive?

Explanation:

Step1: Define Variables

Let \( x \) be the number of \(\$20\) price increases. The price per ticket becomes \( p = 400 + 20x \), and the number of ticket holders becomes \( n = 900 - 15x \).

Step2: Revenue Function

Revenue \( R \) is price times quantity, so \( R(x)=(400 + 20x)(900 - 15x) \). Expand this:
\( R(x)=400\times900-400\times15x + 20x\times900-20x\times15x \)
\( R(x)=360000 - 6000x + 18000x - 300x^{2} \)
\( R(x)=- 300x^{2}+12000x + 360000 \).

Step3: Find Vertex of Quadratic

For a quadratic \( ax^{2}+bx + c \), the vertex (maximum, since \( a=-300<0 \)) is at \( x =-\frac{b}{2a} \). Here, \( a=-300 \), \( b = 12000 \).
\( x=-\frac{12000}{2\times(-300)}=\frac{12000}{600}=20 \).

Step4: Optimal Price

Substitute \( x = 20 \) into price formula: \( p = 400+20\times20=400 + 400=\$800 \).

Step5: Maximum Revenue

Substitute \( x = 20 \) into revenue function:
\( R(20)=-300\times(20)^{2}+12000\times20 + 360000 \)
\( R(20)=-300\times400+240000 + 360000 \)
\( R(20)=-120000+600000=\$480000 \).

Answer:

The price that maximizes revenue is \(\$800\), and the maximum revenue is \(\$480000\).