Question

4) A bus company charges $2 per ticket but wants to raise the price. The daily revenue is modeled by R(x)=-30(x-6)² + 320, where x is the number of $0.15 price increases and R(x) is the revenue in dollars. What should the price of the tickets be for a maximum profit? Hint: don't forget what price where you started.

Answer 1

$2.90

Explanation:

6 × 0.15 = 0.9

2 + 0.9 = 2.9

R(x) = -30(x - 6)² + 320

This equation for the revenue is a quadratic equation model written in the form of completing the square;

This form appears like so:

f(x) = a(x + b)² + c

It is quite useful, especially for this type of question;

Firstly, if a is positive, the quadratic equation will be a u shaped graph when illustrated, if negative, it will be an n shaped graph;

What this means is if a is positive, the vertex of the graph is the lowest point, i.e. the minimum value of f(x), and if a is negative, the vertex will be the highest point, i.e. the maximum value of f(x);

Secondly, the coordinates of the vertex will be:

(-b, c)

So, with regards to the question:

We have -30 in the position of a, the curve is therefore n shaped and the vertex is the highest point (known as the local maximum);

And in place of b and c, we have -6 and 320, so the coordinates of this local maximum are:

(6, 320)

We interpret this like so:

320 is the highest possible value of R(x), which represents revenue, so $320 is the maximum revenue according to this model, and it is achieved when x = 6, i.e. when the price is increased by $0.90 (= 6 × $0.15);

Finally, to get the new ticket price, we add this to the original price to get $2.90.