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

Answer:

$(a)$ $R(20) = 1400$

$(b)$ $Max(x) = 45$

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

Step-by-step 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:

$Max(x) = \frac{-b}{2a}$

Where: $a = -1$     $b = 90$    

So:

$Max(x) = \frac{-90}{2*-1}$

$Max(x) = \frac{-90}{-2}$

$Max(x) = 45$

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