A school typically sells 500 yearbooks each year for $50 each. the economics class does a project and discovers that they can sell 125 more yearbooks for every $5 decrease in price.the revenue for the school sales is R(x)=(500+125x)(50-5x) To maximize profit, what price should the school charge for the yearbooks? What is the possible maximum revenue? If the school attains the maximum revenue, how many yearbooks will they sell?

They should sell each yearbook 5 dollars cheaper because they will make 4,500 dollars more if they did that. With the discount and extra yearboos they will earn $29,250.

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MsEHolt MsEHolt · Answer 2 · 2019-03-21T01:32:07+01:00

Answer:

$35 per yearbook; $30,625

Step-by-step explanation:

In this function, x represents the number of $5 decreases. To find the vertex, we first find the axis of symmetry. We will first write this function in standard form, by multiplying:

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

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

= 25000+-2500x+6250x-625x²

= 25000+3750x-625x²

= -625x²+3750x+25000

The axis of symmetry is found using the formula

x = -b/2a

x = -3750/2(-625) = -3750/(-1250) = 3

Next we use this value of x to find the y-coordinate of the vertex. We substitute this into the function:

R(3) = -625(3²)+3750(3)+25000

= -5265+11250+25000 = 30625

This makes the vertex (3, 30625). This means with 3 $5 decreases, the revenue will be $30,625.

3 $5 decreases means decreasing 3(5) = $15 from the price. This means they sell the yearbooks for 50-15 = $35.