You are selling a lip balm that you created at a local grocery store. Currently, the price of your homemade lip balm is $3.00 per tube. At this price you sell about 300 tubes a month. You are considering changing the price in order to make more money. Obviously, you can't make the price too low, or you will be giving away lip balm. If you make it too high, then people will not buy it. In order to figure out the price you should sell it at in order to make the most money, you do some testing and find that for every $0.10 decrease in price, the number of sales increases by 20 tubes per month.
To determine the price that will maximize your revenue, you need to find the point where the increase in sales resulting from a decrease in price compensates for the lower price per tube. Here's how you can calculate it:
1. Determine the sales increase per $0.10 decrease in price: Since you found that the number of sales increases by 20 tubes per month for every $0.10 decrease in price, divide the increase in sales by the decrease in price: 20 tubes/$0.10 = 200 tubes per dollar.
2. Calculate the additional revenue generated by the increase in sales: Multiply the sales increase per dollar by the current price of $3.00 to find the additional revenue generated by the change in price: 200 tubes per dollar * $3.00 = $600 per dollar.
3. Determine the overall revenue change when you decrease the price: Subtract the revenue decrease resulting from the lower price per tube from the additional revenue generated by the sales increase. As you decrease the price by $0.10, the revenue decrease per tube is $0.10. Therefore, the overall revenue change per tube is $600 per dollar - $0.10 per tube.
4. Evaluate the revenue change per tube for different price changes: To find the optimal price, compare the revenue change per tube for different price decreases. Since we know that the revenue change per tube is $600 per dollar - $0.10 per tube, we can calculate the revenue change per tube for different price decreases:
- For a $0.10 decrease in price: $600 per dollar - $0.10 per tube = $599.90 per tube
- For a $0.20 decrease in price: $600 per dollar - $0.20 per tube = $599.80 per tube
- Continue this calculation for different price decreases until you see a decline in the revenue change per tube.
5. Find the optimal price: The optimal price will occur where the revenue change per tube is highest. Based on the example calculations provided, decreasing the price by $0.10 generates the highest revenue change per tube ($599.90 per tube). Hence, the optimal price to maximize revenue could be $2.90 per tube.
Please note that this analysis assumes a linear relationship between price and sales, which may not hold true in all cases. It is always good to continuously monitor and analyze your sales data to make informed decisions about pricing and revenue optimization.
To determine the price at which you can make the most money, follow these steps:
Step 1: Calculate the revenue at the current price of $3.00 per tube.
Revenue at $3.00 per tube = Price per tube * Number of tubes sold
Revenue at $3.00 per tube = $3.00 * 300 = $900.00
Step 2: Analyze the effect on revenue with a $0.10 decrease in price.
For every $0.10 decrease in price, the number of sales increases by 20 tubes per month.
Step 3: Calculate the new number of tubes sold and revenue with a $0.10 decrease in price.
New number of tubes sold = Current number of tubes sold + Increase in sales
New number of tubes sold = 300 + 20 = 320
New revenue at $2.90 per tube = Price per tube * New number of tubes sold
New revenue at $2.90 per tube = $2.90 * 320 = $928.00
Step 4: Repeat steps 2 and 3, gradually decreasing the price by $0.10 each time and calculating the resulting revenue. Continue until the increase in sales no longer compensates for the decrease in price.
Step 5: Compare the revenue at each price point and choose the price that yields the highest revenue.
Based on the given information, you can now perform the analysis by decreasing the price by $0.10 and calculating the corresponding revenue at each step.