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

Answer:

Explained below.

Step-by-step 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:

$p(x)=400-\frac{1}{10}(x-1000)$

       $=400-\frac{x}{10}+100\\\\=500-\frac{x}{10}$

Thus, the demand function is, $p(x)=500-\frac{x}{10}$.

(b)

The revenue function is:

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

        $=x[500-\frac{x}{10}]\\\\=500x-\frac{x^{2}}{10}$

Maximize the revenue as follows:

      $\frac{d}{dx}(R(x))=0$

$\frac{d}{dx}[500x-\frac{x^{2}}{10}]=0$

        $500-\frac{x}{5}=0$

                 $\frac{x}{5}=500$

                 $x=2500$

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:

$p(2500)=500-\frac{2500}{10}=500-250=250$

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:

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

        $=500x-\frac{x^{2}}{10}- 73000 - 110x\\\\=390x-\frac{x^{2}}{10}- 73000$

Compute the marginal profit as follows:

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

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

$P'(x)=0\\\\390-\frac{x}{5}=0\\\\x=390\times 5\\\\x=1950$

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:

$p(1950)=500-\frac{1950}{10}=500-195=305$

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