Question

An electronics store sells about 40 MP3 players per month for $90 each. For each $5 decrease in price, the store expects to sell 4 more MP3 players.
What value of x gives the maximum monthly revenue?

x =

How much should the store charge per MP3 player to maximize monthly revenue? Round your answer to the nearest dollar.

Answer 1

Step-by-step explanation:

Let's establish our equation first:

• for every $5 decrease, there's an additional of 4 MP3 players sold.
• to get the monthly revenue, we need to multiple the cost of each player to the number of units sold

$\frac{40mp3}{month} \times \frac{90dollars}{mp3} = 3600 \frac{dollars}{month}$

The equation above is for the basis month.

But the next month, we decreased the cost of mp3 player to sold 4 more units.

$\frac{(40 + 4)mp3}{month} \times \frac{(90 - 5)dollars}{mp3} = 3740 \frac{dollars}{month}$

And the next month, we decreased the cost again to gain 4 more additional units sold.

$\frac{(40 + 4x)mp3}{month} \times \frac{(90 - 5x)dollars}{mp3} = revenue per \: month$

If we substitute x with 2, we get 3840.

If we substitute x with 3, we get 3900.

If we substitute x with 4, we get 3920.

If we substitute x with 5, we get 3900.

At fifth time we decreased our price, we also got lesser revenue.

Therefore, our highest revenue would be $3,900.00 per month at our 4th price decrease with a price of $70.00.