03 points l Prenous Armwers scalcCT a 4 7 502.xp My Notes Ask Your Teacher A man
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03 points l Prenous Armwers scalcCT a 4 7 502.xp My Notes Ask Your Teacher A manufacturer has been selling 1000 flat-screen TVs a week at s500 each. A market survey indicates that for each $10 rebate offered to the buyer, the number of Tvs sold will increase by 100 per week. (a) Find the demand function (price as a function of units sold x) p(x) = | 1 (b) How large a rebate should the company offer the buyer in order to maximize its revenue? s 125 (e if its w eekly cost function is C K 64,000 + 100x, how should the manufacturer set the size of the rebate in order to maximize its profit?Explanation / Answer
a)
p(1000) = 450
Sale will increase by $100 if there will be a rebate of $10.
This indicates that for the price decrease of (1 / 100) *10 i.e 1 / 10, the sale will increase by $100.
Therefore,
p(x) = (-1/10) (x) + b
when p(1000) = 450
450 = (-1/10) (1000) + b
450 = -100 + b
b = 550
Therfore, p(x) = (-1/10)*(x) + 550
b)
R(x) = x * p(x)
= x * [(-1/10)*(x) + 550]
= (-1/10)*(x2) + 550x
Taking first derivative
R'(x) = (-1 /5)*x + 550
Equating R'(x) = 0
(-1 /5)*x + 550 = 0
(1/5)*x = 550
x = 550*5
x = 2750
Putting value of x in demnd function
p(2750) = (-1/10) (2750) +550
= 275
Therefore to achieve maximum rebate the price should be set at p(2750) = 275.
Rebate = 450 - 275 = 175
c)
C(x) = 64000 + 100x
Profit function P(x) = R(x) - C(x)
= (-1/10)*(x2) + 550x - (64000 + 100x)
= (-1/10)*(x2) + 450x - 64000
Marginal Profit = P'(x) = (-1/5)*(x) + 450
To maximise profit put P'(x) = 0
(-1/5)*(x) + 450 = 0
(1/5)*(x) = 450
x = 450 *5
x = 2250
Putting value of x in demnd function
p(2250) = (-1/10) (2250) +550
= 325
Therefore to achieve maximum rebate the price should be set at p(2250) = 325
Rebate = 450 - 325 = 125
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