You have been hired as a marketing consultant to Johannesburg Burger Supply, Inc., and you wish to come up with a unit price for its hamburgers in order to maximize its weekly revenue. To make life as simple as possible, you assume that the demand equation for Johannesburg hamburgers has the linear form q = mp + b, where p is the price per hamburger, q is the demand in weekly sales, and m and b are certain constants you must determine.
(a) Your market studies reveal the following sales figures: When the price is set at $2.00 per hamburger, the sales amount to 7000 per week, but when the price is set at $4.00 per hamburger, the sales drop to zero. Use these data to calculate the demand equation.
q =
(b) Now estimate the unit price that maximizes weekly revenue.
$
Predict what the weekly revenue will be at that price.
(a) To calculate the demand equation, we can use the given information about sales figures. We have two sets of data points: when the price is $2.00, the sales amount to 7000 per week, and when the price is $4.00, the sales drop to zero.
Let's write down the demand equation in the given linear form: q = mp + b
Considering the first data point:
- When the price is $2.00, the sales amount to 7000 per week.
So, we have: q = 7000 and p = 2.00.
Plugging in these values into the demand equation, we get: 7000 = 2m + b. (Equation 1)
Considering the second data point:
- When the price is $4.00, the sales drop to zero.
So, we have: q = 0 and p = 4.00.
Plugging in these values into the demand equation, we get: 0 = 4m + b. (Equation 2)
Now, we have a system of two equations (Equation 1 and Equation 2) with two unknowns (m and b). We can solve this system of equations to find the values of m and b.
To eliminate b, we subtract Equation 2 from Equation 1:
(7000 - 0) = (2m + b) - (4m + b)
7000 = 2m - 4m
7000 = -2m
Dividing both sides by -2, we get:
m = -7000 / -2
m = 3500
Substituting the value of m back into either Equation 1 or Equation 2, we can solve for b. Let's use Equation 1:
7000 = 2(3500) + b
7000 = 7000 + b
Subtracting 7000 from both sides, we get:
0 = b
Therefore, the demand equation is q = 3500p.
(b) To estimate the unit price that maximizes weekly revenue, we can use the demand equation q = 3500p and the concept of revenue.
The revenue formula is given by: Revenue = Price × Quantity.
In this case, the revenue formula becomes: R = p × q = p × 3500p = 3500p^2.
To maximize weekly revenue, we need to find the value of p that maximizes the equation 3500p^2.
This can be done by finding the critical point of the quadratic equation. To find the critical point, we can take the derivative of the revenue equation with respect to p and set it equal to zero:
dR/dp = 2 × 3500p = 0
Simplifying, we get:
7000p = 0
Dividing both sides by 7000, we find:
p = 0
So, the unit price that maximizes weekly revenue is $0.
However, since a price of $0 is not practical, it is likely that the given information is not sufficient to determine the unit price that maximizes weekly revenue. Additional data or analysis would be required to find a more reasonable and practical unit price.
Without a specific unit price, we cannot accurately predict the weekly revenue at that price.