Question

A school typically sells 500 yearbooks in a year for $50 each. The economics class does a projectand discovers that they can sell 125 more yearbooks for every $5 decrease in price, The revenue foryearbook sales is R(x) = (500 + 125x)(50 - 5x).

Answer 1

Part 1. Given the equation

$R(x)=(500+125x)(50-5x)$

Simplify the equation

$\begin{gathered} R(x)=500\cdot50+500(-5x)+125x\cdot50+125x(-5x) \\ R(x)=25000-2500x+6250x-625x^2 \\ R(x)=25000+3750x-625x^2 \end{gathered}$

Then, we differentiate the equation and we equal to zero

$dR(x)=3750-2\cdot625x=3750-1250x$

dR(x)=0

$\begin{gathered} 3750-1250x=0 \\ 3750-1250x-3750=0-3750 \\ -1250x=-3750 \\ \frac{-1250x}{-1250}=\frac{-3750}{-1250} \\ x=3 \end{gathered}$

Therefore The price is equal to:

$50-5x=50-5(3)=50-15=35$

Answer: $35

Part 2. The possible maximum revenue is

$\begin{gathered} R(x)=25000+3750(3)-625(3)^2 \\ R(x)=25000+11250-5625=30625 \end{gathered}$

Answer: $30625

Part 3. The number of yearbooks sold:

$500+125x=500+125(3)=500+375=875$

Answer: 875 yearbooks