Explanation:
To find the revenue-maximizing price and the maximum revenue, you can follow these steps:
a) Write a function to represent the revenue from sales:
Let's define the variables:
x: the price decrease in dollars (for example, if the original price is $6.00, and x = 0.50, the new price is $5.50).
S(x): the number of additional smoothies sold when the price decreases by x dollars.
You are given that for every $0.50 decrease in price, the store will sell 50 additional smoothies. Therefore, you can express S(x) as:
S(x) = 50x
Now, the total number of smoothies sold when the price is reduced by x dollars will be 200 (the original number) plus the additional smoothies:
Total Smoothies Sold = 200 + S(x) = 200 + 50x
The revenue (R) can be calculated by multiplying the price per smoothie (P) by the total number of smoothies sold:
R(x) = P(x) * Total Smoothies Sold
R(x) = (6.00 - x) * (200 + 50x)
This is the function in factored form to represent the revenue from sales.
b) Determine the number of smoothies they must sell to maximize their revenue:
To find the number of smoothies that maximize revenue, you need to find the value of x that maximizes R(x) by taking the derivative of R(x) with respect to x and setting it equal to zero:
dR(x)/dx = 0
c) Determine the maximum revenue that can be achieved:
Once you've found the value of x that maximizes revenue in part (b), plug that value back into the revenue function R(x) to find the maximum revenue.
Let's proceed with part (b) and (c):
b) To find the value of x that maximizes revenue, take the derivative of R(x) and set it equal to zero:
dR(x)/dx = 0
R(x) = (6.00 - x) * (200 + 50x)
Now, calculate the derivative:
dR(x)/dx = (200 + 50x) * (-1) + (6.00 - x) * 50
Set it equal to zero and solve for x:
(200 + 50x) * (-1) + (6.00 - x) * 50 = 0
Now, solve for x.
c) Once you've found the value of x that maximizes revenue, plug that value back into the revenue function R(x) to find the maximum revenue.