A store has been selling 300 Blu-ray disc players a week at $600 each. A market survey indicates that for each $40 rebate offered to buyers, the number of units sold will increase by 80 a week. Find the demand function and the revenue function. How large a rebate should the store offer to maximize\

75 $

Step-by-step explanation:

According to problem statement p(300) = 600

And we know that with a rebate of 40 $, numbers of units sold will increase by 80 then if x is number of units sold, the increase in units is

(  x  - 300 )  , and the price decrease

(1/80)*40  =  0,5

Then the demand function is:

D(x)  =  600  - 0,5* ( x - 300 )  (1)

And revenue function is:

R(x) =  x * (D(x)   ⇒   R(x) =  x* [  600  - 0,5* ( x - 300 )]

R(x) = 600*x  - 0,5*x * ( x - 300 )

R(x) = 600*x - 0,5*x² - 150*x

R(x) = 450*x  - (1/2)*x²

Now taking derivatives on both sides of the equation we get

R´(x) =  450  - x

R´(x) =  0       ⇒   450  - x = 0

x = 450 units

We can observe that for   0 < x  < 450  R(x) > 0 then R(x) has a maximum for x = 450

Plugging this value in demand equation, we get the rebate for maximize revenue

D(450)  =  600  - 0,5* ( x - 300 )

D(450)  =  600 - 225 + 150

D(450)  =

D(450)  =  600 - 0,5*( 150)

D(450)  =  600 - 75

D(450)  = 525

And the rebate must be

600 - 525  = 75 $

  y= -3x is the required function.

step-by-step explanation:

we have been given a table:

x             y

0             0

1             -3

3             -9

-2             6

we can see that the values of y is -3 times the value of x

hence, the function would be: y= -3x.

put the values of x will give the same y as in the table that means function is satisfying the given table.

3; 9 is 8 to -4

step-by-step explanation: