EXAMPLE 6 A store has been selling 300 DVD burners a week at $400 each. A market survey indicates that for each $30 rebate offered to buyers, the number of units sold will increase by 60 a week. Find the demand function and the revenue function. How large a rebate should the store offer to maximize its revenue?

See below

Step-by-step explanation:

Let x be the number of $30 rebates performed on the product.

Since that for each $30 rebate offered to buyers the number of units sold will increase by 60 a week, the demand function D is given by

D(x)= 300 + 60x units per week

The revenue R is given by the amount of units sold multiplied by the price of each unit  

R(x)= (300 + 60x)(400 - 30x) dollars

We must now find a value of x that maximizes the revenue R.

Taking the first derivative using the derivative of a product, we get

R'(x) = (300 + 60x)'(400 - 30x) + (300 + 60x)(400 - 30x)' =

= 60(400 - 30x) + (300 + 60x)(-30) = 15000 - 3600x

The critical point is obtained when R'(x) = 0,

that is, when 3600x = 15000 or x = 15000/3600 = 4.167

So, the critical point is x = 4.167.

Taking the second derivative

R''(x) = -3600 < 0, so x = 4.167 is a maximum.

Since the number $30 rebates should be an integer, we take the closest integer to 4.167 which is 4.

The number of $30 rebates that maximizes the revenue is 4.

c

step-by-step explanation:

for this case we must apply distributive property to the following expression:

the distributive property states that:

then, when applying the property we have to multiply the -2 for each term within the parenthesis: