Question

The revenue R in dollars received from selling radios is () = (90 − ). a) Evaluate (20) and interpret the result. b) What number of radios sold maximize revenue? c) What number of radios should be sold for revenue to be $2000 or more?

Answer 1

`(a)`
`R(20) = 1400`

`(b)`
`Max(x) = 45`

`(c)`
`x = 40` or
`x = 50`

Explanation:

Given

`R(x) = x(90 - x)`

Solving (a): R(20)

Substitute 20 for x

`R(20) = 20 * (90 - 20)`

`R(20) = 20 * 70`

`R(20) = 1400`

Interpret: The cost of 20 radios is $1400

Solving (b): Maximum radio that maximizes the function

`R(x) = x(90 - x)`

Open bracket

`R(x)=90x - x^2`

`R(x)=- x^2 + 90x`

To get the maximum, we make use of:

`Max(x) = (-b)/(2a)`

Where:
`a = -1`
`b = 90`

So:

`Max(x) = (-90)/(2*-1)`

`Max(x) = (-90)/(-2)`

`Max(x) = 45`

The number of radio that maximizes the function is 45

Solving (c): Number of radios that amounts to $2000

`R(x) = x(90 - x)`

Substitute 2000 for R(x)

`2000 = x(90 - x)`

Open bracket

`2000 = 90x - x^2`

Express as quadratic

`x^2 - 90x + 2000 = 0`

Expand

`x^2 - 40x - 50x + 2000 = 0`

Factorize:

`x(x - 40) - 50(x - 40) = 0`

`(x - 40)(x - 50) = 0`

This gives

`x = 40` or
`x = 50`

40 or 50 items when sold will amount to $2000