A manufacture has been selling 1850 television sets a week at $390 each. A marke

Question

A manufacture has been selling 1850 television sets a week at $390 each. A market survey indicates that for each $16 rebate offered to a buyer, the number of sets sold will increase by 160 per week.

1) Find the price p(x) of each television as a function of x, where x is the number of the television sets sold per week.
p(x)=

2) How large a rebate should the company offer, in order to maximize its revenue? dollars

3) If the weekly cost function is 120250+130x , what value of the rebate maximizes the profit? dollars

Explanation / Answer

1)let p(x) = kx +y ,
where k&y are two constants.
390 = 1850k +y   eq(1)
390-16 =(1850+160)k +y
374 = 2010k+y   eq(2)
substract eq(1) from eq(2)
160k = -16
k = -16/160 =-1/10
y =390 -1850k= 390 -(-185) =390+185 =575
now
p(x) = -x/10 +575
p(x) =575 -(x/10)

2)revenue R = p(x)*x =575x - (x^2)/10
R' =575 -(x/5)
for maximum of R , R' =0
575-(x/5) =0
575=x/5
x =575*5 =2875
p(2875) =575- 2875/10 =575-287.5 =287.5
rebate =390 -287.5 = 102.5
rebate should the company offer to a buyer, in order to maximize its revenue =$102.5

3)

profit = x(575 - x/10) - (120250+130x) .......... revenue - cost
dprofit/dx = 575 - x/5-130
445= x/5 ......... set dprofit/dx = 0 to find the stationary points
x = 2225
p = 575 - 2225/10 = 575-222.5=352.5
390-352.5 = $37.5

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