Before we can begin solving for the price that will generate the maximum revenue, we need to create an equation for the revenue using the given information. Let's look for the key words in the given text.
Revenue is the product of the number of phones sold and the price of the phones
⇓
r = n * p
We are also told that the number of phones increases by 20 while the price decreases by 1. Let's make a table of values with the corresponding information.
Number of Phones (n)
Price (p)
Revenue (np)
500
75
37500
520
74
38480
540
73
39420
560
72
40320
Next, we can introduce a variable x to write a general formula for finding the revenue based on the number of phones sold.
Counter
Number of Phones (n)
Price (p)
Revenue (np)
0
500 + 20 ( 0) = 500
75- 0 =75
37500
1
500+ 20( 1) = 520
75 - 1 = 74
48480
2
500+ 20( 2) = 540
75 - 2 = 73
39420
3
500 + 20( 3) = 560
75 - 3 = 72
40320
...
...
...
...
x
500 + 20 x
75 - x
(500+ 20 x)(75 - x)
Now, we can use the generalization for revenue as r(x)=(500+20x)(75-x). To find the maximum revenue we need the function in standard quadratic form, y= ax^2+ bx+ c.
Now, we can use standard form to find the vertex of this function, which will give us the maximum revenue value of x. Let's find the axis of symmetry.
x=- b/2 a
In our case, we can substitute a= -20 and b= 1000 into the formula for the axis of symmetry.
The axis of symmetry gives us the x-coordinate for vertex, x= 25. Finally, to find the price at this value of x, we can use our formula for price.
75 - x=p ⇒ 75- 25=50
The phone price that should be used to gain the maximum revenue is $50.
