A street vendor sells about 20 shirts each day when she charges $8 per shirt. If she decreases the price by $2, she sells about 10 more shirtseach day. How can she maximize daily revenue?
To maximize daily revenue, the street vendor needs to determine the optimal price at which she can sell the maximum number of shirts each day.
Let's break down the given information:
1. When the price is $8 per shirt, she sells about 20 shirts per day.
2. When the price is decreased by $2, she sells about 10 more shirts per day.
To find the optimal price, we can start by examining the relationship between the price and the number of shirts sold.
Price ($8) --> Shirts Sold (20)
Price ($6) --> Shirts Sold (30)
As you can see, decreasing the price by $2 leads to an increase in shirt sales by 10.
Now, let's continue this process to find more data points:
Price ($8) --> Shirts Sold (20)
Price ($6) --> Shirts Sold (30)
Price ($4) --> Shirts Sold (40)
Price ($8) --> Shirts Sold (20)
Price ($10) --> Shirts Sold (10)
From the data above, we can observe that decreasing the price by $2 increases the number of shirts sold, while increasing the price by $2 decreases the number of shirts sold.
To maximize daily revenue, the vendor needs to find the price point at which the increase in the number of shirts sold compensates for the decrease in price.
Since decreasing the price by $2 leads to 10 more shirts sold, and each shirt generates $6 of revenue ($8 - $2), the vendor's revenue increases by $60 (10 shirts x $6) by decreasing the price.
However, increasing the price by $2 decreases the number of shirts sold by 10, resulting in a decrease of $80 in revenue (10 shirts x $8).
Therefore, in order to maximize daily revenue, the vendor should lower the price by $2.