A manufacturer has been selling 1000 television sets a week at $400 each. A mark

Question

A manufacturer has been selling 1000 television sets a week at $400 each. A market survey indicates that for each $10 rebate offered to the buyer, the number of sets sold will increase by 100 per week. Round your answers to the nearest dollar.

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

p(x) =

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

$

(c) If the company experiences a cost of C(x) = 68,000 + 130x, how should the manufacturer set the size of the rebate in order to maximize its profit?

$

An airline finds that if it prices a cross-country ticket at $200, it will sell 300 tickets per day. It estimates that each $10 price reduction will result in 50 more tickets sold per day. Find the ticket price (and the number of tickets sold) that will maximize the airline's revenue.

ticket price   $     

number of tickets           

Find the point of diminishing returns for the profit function where x is the amount spent on marketing in million dollars.

P(x) = 4x + 30x2 2.5x3

for

0 x 6

million dollars

Explanation / Answer

a) P(x)= -x10/100+k

P(x)=-x/10+k

400=-100+k

k=500

P(x)= -x/10+500

b) R(x)= -(x^2)/10+500x

R'(x) = -x/5+500

R'(x)=0 for maximum revenue

x=2500

P(2500) = -250+500=250

A rebate of 150 to maximize revenue.

c)Pr(x)= R(x)-C(x)=-(x^2)/10+500x-68,000 - 130x

Pr'(x)= -x/5+500-130= -x/5+370

for maximum profit Pr'(x)=0

x=1850

P(1850) = -185+500=315

So a rebate of 85 will maximize profit

Problem 2

p(x)= -x/5+k

200=-60+k

k=260

p(x)=-x/5+260

R(x)= -x^2/5+260x

R'(x)= -2x/5+260

For maximum revenue R'(x)=0

x=650     ticket sold

p(x)=-x/5+260

p(650)= -130+260=130   ticket price

Problem 3

P(x) = 4x + 30x2 2.5x3

P'(x) = 4+60x-7.5x^2

P''(x)=60-15x

For point of diminishing returns

P''(x)=0

x=4

p(4) = 16+480-160=336 million dollars

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