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.

answer: the value should be used for i in the formula of calculation the monthly payment , the interest rate per period is 0.85.

step-by-step explanation:

the given rate of interest i= 85%

in calculating the monthly payment for a loan with an interest rate compounded monthly, the value of the interest rate per period we use in the formula must be in decimal form.

to convert it in decimal, we divide it by 100, we get

hence, the value should be used for i in the formula of calculation the monthly payment , the interest rate per period is 0.85.