A manufacturer has been selling 1000 flat-screen TVs a week at $450 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.

Question

(a) Find the demand function (price p as a function of units sold x).

p(x)=______

(b) How large a rebate should the company offer the buyer in order to maximize its revenue?

$=_____

(c) If its weekly cost function is C(x) = 78,000 + 130x, how should the manufacturer set the size of the rebate in order to maximize its profit?

$=______

Answer 1

Assuming the demand Q(p), is LINEAR function of the price...

Q(450) = 1000

dQ/dp = -100/10 = -10 <--- the slope of the demand line is ALWAYS negative

point-slope form is:

Q-1000 = -10(p-450)

Q-1000 = -10p + 4500

Q = -10p + 5500

So Q(p) = -10p +5500 <-- will need this function later

The inverse is

(Q-5500)/-10 = p

(-1/10)Q + 550 = p

So p(x) = (-1/10)x + 550 where Q=x

is the answer to part A

(b) the REVENUE is a function of the price,

defined as Revenue = price * demand

or REvenue = price * qty sold

That is,

R(p) = p*Q(p)

= p * (-10p + 5500)

= -10p^2 + 5500p

maximizing:

0 = dR/dp = -20p + 5500

-5500 = -20p

p=-5500/-20

p = 550/2 = 275

So the max. revenue

-10 (275)^2 + 5500 *275

= -756250 + 1512500

= 756250

(c) C(x) is the cost to produce x television sets

C(Q(p)) is the cost to produce the demanded quantity

C(q(p)) = 78000 + 130*Q(p)

= 78000 + 130* (-10p +5500)

= 78000 + -1300p + 715000

= 793000 - 1300p

Profit = Revenue - cost, where Profit,P, is a function of price,p.

P(p) = -10p^2 + 5500p - ( 793000 - 1300p)

= -10p^2 + 5500p - 793000 + 1300p

= -10p^2 + 6800p - 793000

Maximizing:

0=dP/dp = -20p + 6800

so max profit occurs at price p=-6800/-20 = $340