A bicycle manufacturing company knows that the equation R=-2.5p?+ 500 p models the relationship between the price (p) and the revenue (R) for one of its product lines.
What is the maximum revenue possible?
$

The maximum revenue is $25,000

Step-by-step explanation:

The model equation is;

R = -2.5p^2 + 500p

To find the maximum value, we differentiate this and set result = 0

The first differential of this model equation is -5p + 500

we set this to zero to get the maximum price

5p = 500

p= 500/5

p = 100

Now to get the maximum revenue, we simply substitute the value of the maximum price

That would be -2.5(100)^2 + 500(100)

= -2.5(10,000) + 50,000

= 25,000 + 50,000

= 25,000

step-by-step explanation:

you divide the denominator by the numerator on both sides and if the quotients are equal then the fractions are equivalent

the revenue of the fitness center, y, is the product of two factors: the number of members and the fee paid by each member. there are 320 members now, but number of members goes down by 10 for each $5 increase in fees, so if x is the number of $5 price increases then members is equal to (320 - 10x). since the fee is currently $45, and the fee increases by $5 for every price bump x, the fee is given by (45 + 5x). so, revenue y is given by y = (320 - 10x)(45 + 5x).