Answer:
The price to yield a maximum revenue = $52 per unit
Explanation:
Given Data:
Selling price = $30 per unit
Unit per week = 296 units
Price increase = $5
Unit dropped = 276 units
For demand function;
D = ax +b ------------------------1
when x = $30 and D = 296, equation 1 becomes
30a + b = 296 ----------------2
When price increases by $5, x = 30+5 = $35 and D = 276
Equation 1 becomes,
35a + b = 276 -------------------3
Solving equation 2 and 3 simultaneously to obtain a and b, we have
Making b subject in equation 2-
b = 296 - 30a ---------------4
putting equation 4 into equation 3, we have
35a + 296 - 30a = 276
5a = -20
a = -4
Putting the value for a into equation 4, we have
b = 296 -30(-4)
= 296 +120
= 416
Equation 1 becomes;
D = -4x + 416
revenue function is given as;
R(x) = D*x
= (-4x + 416)x
= -4x² + 416x
For maximum revenue, dR(x)/dx = 0
dR(x)/dx = -4x² + 416x = 0
-8x + 416 = 0
x = 0 - 416/-8
x = -416/-8
x = $52
Therefore, the price to yield a maximum revenue = $52 per unit