The manager of a home electronics store is considering repricing a new model of digital camera. At the current price of $680 the store sells about 70 cameras each month. Sales data from other stores indicate that for each $20 decrease in price about 5 more cameras per month would be sold. How much should the manager charge for a camera to maximize monthly revenue? What is the maximum revenue.?
To determine the optimal price that maximizes monthly revenue, we need to analyze the relationship between price, quantity sold, and revenue. Let's break it down step by step:
1. Start by determining the demand function, which represents the relationship between price and quantity sold. Based on the given information, we know that for each $20 decrease in price, 5 more cameras are sold per month. Therefore, we can write the demand function as:
Quantity Sold = 70 + (5 * ((Current Price - New Price) / $20))
In this case, the current price is $680, so the demand function becomes:
Quantity Sold = 70 + (5 * ((680 - New Price) / $20))
2. Next, we can calculate the monthly revenue, which is the product of price and quantity sold:
Revenue = Price * Quantity Sold
3. To find the price that maximizes revenue, we can express the revenue function in terms of the price variable and then differentiate it with respect to price to find its critical points.
Revenue = Price * (70 + (5 * ((680 - Price) / $20)))
Revenue = Price * (70 + (5 * (34 - (Price / $20))))
4. Differentiating the revenue function with respect to price:
Revenue' = (70 + (5 * (34 - (Price / $20)))) + (Price * 5 * (-1 / $20))
Revenue' = 70 + (170 - (6.5 * Price)) - (Price / 4)
Revenue' = 240 - (5.5 * Price)
5. Set the derivative equal to zero and solve for Price:
240 - (5.5 * Price) = 0
5.5 * Price = 240
Price = $43.64 (rounded to the nearest cent)
6. To confirm that this is indeed a maximum, we can take the second derivative of the revenue function and evaluate it at the critical point:
Revenue'' = -5.5 (the second derivative is negative, indicating a maximum)
7. Now that we know the optimal price is $43.64, we can substitute it back into the demand function to find the quantity sold:
Quantity Sold = 70 + (5 * ((680 - 43.64) / $20))
Quantity Sold ≈ 97 cameras
8. To calculate the maximum revenue, substitute the optimal price and quantity into the revenue function:
Revenue = $43.64 * 97
Revenue ≈ $4,233.08
Therefore, the manager should charge $43.64 for the camera to maximize monthly revenue, and the maximum revenue achievable is approximately $4,233.08.