Question

A manufacturer has been selling 1000 flat-screen TVs a week at $400 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 p as a function of units sold 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) = 73,000 + 110x, how should the manufacturer set the size of the rebate in order to maximize its profit?

Answer 1

Explained below.

Explanation:

(a)

It is provided that the price function for 1000 TVs is,

p (1000) = 400.

Also provided that if rebate of $10 is given then sale increases by 100 per week.

Let x be the number of unit sold per week then (x − 1000) is the increase in the number of units sold.

Then the price function is:

`p(x)=400-(1)/(10)(x-1000)`

`=400-(x)/(10)+100\\\\=500-(x)/(10)`

Thus, the demand function is,
`p(x)=500-(x)/(10)`.

(b)

The revenue function is:

`R(x)=x\cdot p(x)`

`=x[500-(x)/(10)]\\\\=500x-(x^(2))/(10)`

Maximize the revenue as follows:

`(d)/(dx)(R(x))=0`

`(d)/(dx)[500x-(x^(2))/(10)]=0`

`500-(x)/(5)=0`

`(x)/(5)=500`

`x=2500`

Observe that R'(x) > 0 for 0 ≤ x < 2500 and R'(x) < 0 for x > 2500. Hence, first derivative test will lead to the conclusion that maximum occurs at x = 2500.

Compute the value p (2500) as follows:

`p(2500)=500-(2500)/(10)=500-250=250`

Then the rebate to maximize the revenue should be: $400 - $250 = $150.

(c)

The weekly cost function is,

`C(x) = 73000 + 110x`

Compute the profit function as follows:

`P(x)=R(x)-C(x)`

`=500x-(x^(2))/(10)- 73000 - 110x\\\\=390x-(x^(2))/(10)- 73000`

Compute the marginal profit as follows:

`\text{Marginal profit}=P'(x)\\=(d)/(dx)P(x)\\=(d)/(dx)[390x-(x^(2))/(10)- 73000]\\=390-(x)/(5)`

Compute the value of x for P'(x) = 0 as follows:

`P'(x)=0\\\\390-(x)/(5)=0\\\\x=390* 5\\\\x=1950`

Observe that P'(x) > 0 for 0 ≤ x < 1950 and P'(x) < 0 for x > 1950. Hence, first

derivative test will lead to the conclusion that maximum occurs at x = 1950.

Compute the value p (1950) as follows:

`p(1950)=500-(1950)/(10)=500-195=305`

Then the rebate to maximize the profit should be: $400 - $305 = $95.