Answer:
[tex]\textsf{a)} \quad P=-20(x-2)^2+3380[/tex]
b) cost of ticket = $13
max profit = $3380
number of tickets sold = 260
Step-by-step explanation:
Profit equation: [tex]P=20(15-x)(11+x)[/tex]
Part (a)
Vertex form of quadratic equation: [tex]y=a(x-h)^2+k[/tex]
(where (h, k) is the vertex and [tex]a[/tex] is some constant)
First, write the given profit equation in standard form [tex]ax^2+bx+c[/tex]:
[tex]\begin{aligned}\implies P&=20(15-x)(11+x)\\& = 20(165+4x-x^2)\\& = -20x^2+80x+3300\end{aligned}[/tex]
Factor -20 from the first two terms:
[tex]\implies P=-20(x^2-4x)+3300[/tex]
Complete the square:
[tex]\implies P=-20(x^2-4x+4)+80+3300[/tex]
[tex]\implies P=-20(x-2)^2+3380[/tex]
Part (b)
The vertex of the profit equation is (2, 3380).
Therefore, the cost of the ticket is when x = 2 ⇒ $11 + 2 = $13
The maximum profit is the y-value of the vertex: $3380
The number of tickets sold at this price is:
⇒ max profit ÷ ticket price
= 3380 ÷ 13
= 260
