Answer:
a) q = -3500p + 14000
b) p=1.8666 ≅ $1.87 is the price that maximizes the revenue.
c) $13,937.78
Step-by-step explanation:
(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.
If we assume that the demand equation has the linear form
q = mp + b
where p is the price per hamburger, q is the demand in weekly sales then we have the following 2 linear equations:
for price p=2
(1) 7000 = 2m +b
for price p=4
(2) 0 = 4m +b
Multiplying equation (1) by -1 and adding it to equation (2)
-7000 + 0 = -2m + 4m -b + b
and
-7000 = 2m, hence m =- 7000/2 = -3500.
Substitute this value of m in (2) to obtain
0 = 4(-3500) + b, hence b = 14000
and the formula for the demand equation is
q = -3500p + 14000
(b) Now estimate the unit price that maximizes weekly revenue.
Let R(p) the function which gives the revenue. Then
Taking the<em> derivative</em> with respect to p, we get
R'(p) = -7500p + 14000
and R'(p) = 0 when p = 14000/7500 = 1.8666
Since R''(1.8666) < 0 then <em>p=1.8666 is a maximum</em>.
Predict what the weekly revenue will be at that price.
The maximum revenue could be estimated as