SOLUTION: 2. The captain of a riverboat charges $36 per person including lunch. The cruise averages 300 customers a day. The captain is considering increasing the price. A survey of the

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Question 321938: 2. The captain of a riverboat charges $36 per person including lunch. The cruise averages 300 customers a day. The captain is considering increasing the price. A survey of the customers indicates that for every $2 increase, there would be 10 fewer customers. What increase in price would maximize the revenue?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
Cost is $36 per person including lunch.

Cruise averages 300 customers a day.

Current Revenue is modeled by R = 36 * 300

Every $2 increase in price yields 10 fewer customers.

If you let x = change in price and change in number of customer, then you get an equation that looks like this:

R = (36 + 2x) * (300 - 10*x)

When x = 1, the price per person = 38 and the number of customers = 290.
When x = 2, the price per person = 40 and the number of customers = 280.
etc.

By allowing x to represent a unit of change, you can now model what happens to the revenue based on a change in x.

The graph of this equation would look like this:

graph%28600%2C600%2C-25%2C35%2C-15000%2C15000%2C%2836%2B2%2Ax%29+%2A+%28300-10%2Ax%29%29

Formula to model this would be:

R = Total Revenue
P = Price per person
N = number of people.

Currently, P = 36 and N = 300 which makes R = 36 * 300 = 10800.

On the graph, that is the value of y when x = 0.

When x = 0, there is no change in the current price per person and the number of customers.

You can see from the graph that this curve will peak somewhere between x = 5 and x = 8.

This formula looks like a quadratic equation, so we can use the min/max formula of x = -b/2a to find the min/max point of this equation.

To do that, we need to transform this equation into the standard form of a quadratic equation which is:

ax^2 + bx + c = 0

Our equation is currently in the form of:

R = ((36 + 2x) * (300-10*x))

On the graph, R is represented by y, so we change the formula to show as:

y = ((36 + 2x) * (300-10*x))

We simplify this equation by using the distributive property of multiplication to get:

y = 36 * (300 - 10*x) + 2x * (300 - 10*x)

we simplify further by performing the multiplications indicated to get:

y = 10800 - 360*x + 600*x - 20*x^2

We combine like terms to get:

y = 10800 + 240*x - 20*x^2

We set y = 0 to get:

0 = 10800 + 240*x - 20*x^2

We re-order the terms to get:

0 = -20*x^2 + 240*x + 10800

Since a = b implies that b = a, this equation can be rewritten as:

-20*x^2 + 240*x + 10800 = 0

The equation is now in standard form of a quadratic equation of:

ax^2 + bx + c and we get:

a = -20
b = 240
c = 10800

The x value of the max/min point of this quadratic equation is found by the formula of:

x = -b/2a which becomes:

x = -240 / -40 = 6

The x value of the max / min point is equal to 6.

The y value of the max / min point is equal to f(6).

Since y = f(x) = -20*x^2 + 240*x + 10800, then y = f(6) is equal to -20*(6)^2 + 240*(6) + 10800 which becomes:

y = -20*36 + 240*6 + 10800 which becomes:

y = -720 + 1440 + 10800 which becomes:

y = 11520

I'll repeat the graph with the addition of a horizontal line at y = 11520 so you can see that the maximum point of the graph is there.



The key was in setting up the equation on a 2 dimensional graph so that the value of y represented the revenue and the value of x represented both the increase in price and the drop in number of customers.

The only way I could figure to do this was to allow 2*x to equal the increase in price and 10*x equal the drop in numbers of customers.

y represents the value of the revenue.

x represents the value of a unit of change.

Once the formula was generated, the rest was fairly straightforward.

You just needed to know the standard form of a quadratic equation and the fact that the x value of the max / min point of a parabola (represented by a quadratic equation) was equal to -b/2a.