Answer:
a) The linear function is
P(x) = 600 - x/10
b) $200
c) $145
Step-by-step explanation:
The manufacturer sold a total number of 1000 television in a week
The selling price of each television is $500
The sale increases by 100 per week if a rebate of $10 dollars is offered to the buyer.
Therefore, we have a price decrease of 1/100 * 10 = 1/10 for each unit
Let x be the number sold per week.
x-1000 gives the increase in sales
Let P(x) be the price
P(x) = 500 - 1/10(x - 1000)
= 500 - x/10 + 1000/10
= 500 - x/10 + 100
Collect like terms
P(x) = 500 +100 - x/10
= 600 - x/10
The linear demand function with price as a function of units sold is given as
P(x) = 600 - x/10
b) Let R(x) be the revenue
Revenue = number sold * Price
R(x) = x* P(x)
R(x) = x(600 -x/10)
= 600x - x^2/10
Differentiate R(x) with respect to y
R'(x) = 600 - 2x/10
= 600 - x/5
The revenue is maximum when R(x) = 0
600 - x/5 = 0
600 = x/5
x = 600*5
x = 3000
Remember that P(x) = 600 - x/10
P(3000) = 600 - 3000/10
= 600 - 300
= 300
The rebate to maximize the revenue will be 500 - 300 = $200
c) C(x) = 76000 + 110x
C(x) = Cost
P(x) = R(x) - C(x)
P(x) = (600x - x^2/10) - (76000 + 110x)
= 600x - x^2/10 - 76000 - 110x
Collect like terms
P(x) = 600x - 110x - x^2/10 - 76000
= 490x - x^2/10 - 76000
Differentiate P(x) with respect to x
P'(x) = 490 - 2x/10
= 490 - x/5
P(x) is maximized when it is equal to 0
490 - x/5 = 0
490 = x/5
x = 490*5
x = 2450
P(x) = 600 - x/10
P(2450) = 600 - 2450/10
= 600 -245
= 355
The rebate to maximize the revenue will be 500 - 355 = $145
