Answer:
a) P(x) = 550 - x/10
b) $175
c) $100
Step-by-step explanation:
Number of television sold per week = 1000
The selling price for each = $450
a) If $10 rebate offered to the buyer increases the sale by 100 per week, the price will decrease by
1/100*10 = 1/10 per unit
Let X be the number of television sold per week
The increase in number sold per week = x - 1000
P(x) = 450 - 1/10(x-1000)
P(x) = 450 - x/10 + 100
= 550 - x/10
The demand function is
P(x) = 550 - x/10
b) Let R(x) be the revenue
R(x) = xP(x)
= x[550 - x/10]
= 550x - x^2/10
Differentiate R(x) with respect to x
R'(x) = 550 - 2x/10
= 550 - x/5
R'(x) is the marginal revenue
The revenue is maximized when R'(x) = 0
550 - x/5 = 0
550 = x/5
x = 550*5
x = 2750
Recall that P(x) = 550 - x/10
Put x= 2750
P(2750) = 550 - 2750/10
= 550 - 275
= 275
The rebate to maximize the revenue will be 450 - 275 = $175
c) C(x) = 74000 + 150x
P(x) = R(x) - C(x)
P(x) = 550x - x^2/10 - (74000 + 150x)
= 550x - x^2/10 - 74000 - 150x
Collect like terms
P(x) = 550x - 150x - x^2/10 - 74000
= 400x - x^2/10 - 74000
Differentiate P(x) with respect to x
P'(x) = 400 - 2x/10
= 400 - x/5
P'(x) is the marginal profit.
The profit is maximum when P'(x) = 0
400 - x/5 = 0
400 = x/5
x = 400*5
x = 2000
Recall that P(x) = 550 - x/10
P(2000) = 550 - 2000/10
= 550 - 200
= 350
The rebate to maximize the revenue = 450 - 350
= $100
