A manufacture has been selling 1800 television sets a week at $510 each. A market survey indicates that for each $22 rebate offered to a buyer, the number of sets sold will increase by 220 per week.

Question

A manufacturer has been selling 1800 television sets a week at $510 each. A market survey indicates that for each $22 rebate offered to a buyer, the number of sets sold will increase by 220 per week.

a) Find the function representing the demand p (x), where x is the number of television sets sold per week and p(x) is the corresponding price.

p(x)=___________

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

_____________ dollars.

c) if the weekly cost function is 153000+170x, show if should it set the size of the rebate to maximize its profit? _____________ dollars

Answer 1

a)

If x is the number of units sold, then the increase in sales is:

increase = x-1800

For each increase of 220 units, the price is decreased by $22. This gives a linear equation where the slope,

m = 22/220

So

p(x) = 510-(22/220)(x-1800) = 510-x/10+810 = 690-x/10

b)

Revenue = R(x) = xp(x) = 690x-x²/10

To maximize, find x when R'(x)=0

R'(x) = 690 -x/5 --> x = 3450

You can use the second derivative test to show that this is the maximum.

New Price:

p(3450) = 690-3450/10 = 345

Rebate = 510-345 = $165

c)

If cost, C(x) = 153000+170x

Profit, P(x) = R(x)-C(x)

Marginal profit, P'(x)=R'(x)-C'(x) = 690-x/5-170.

Maximum x = 2600

p(2600) = 690-2600/10 = 430

The maximum profit rebate will be:

Rebate = 510-430 = $80

Answer 2

R(x) = (1800 + 220x)(510 - 22x)

Can you convert to vertext form and answer?