Question

couldddddddddd i have some help super fast?

Answer 1

Answer:

$\textsf{a)} \quad P=-20(x-2)^2+3380$

b)  cost of ticket = $13

    max profit = $3380

    number of tickets sold = 260

Step-by-step explanation:

Profit equation:   $P=20(15-x)(11+x)$

Part (a)

Vertex form of quadratic equation:  $y=a(x-h)^2+k$

(where (h, k) is the vertex and $a$ is some constant)

First, write the given profit equation in standard form $ax^2+bx+c$:

$\begin{aligned}\implies P&=20(15-x)(11+x)\\& = 20(165+4x-x^2)\\& = -20x^2+80x+3300\end{aligned}$

Factor -20 from the first two terms:

$\implies P=-20(x^2-4x)+3300$

Complete the square:

$\implies P=-20(x^2-4x+4)+80+3300$

$\implies P=-20(x-2)^2+3380$

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