A Company sells a particular model of bicycle for (x) dollars and predicts that the revenue from selling the bicycles is given by f(x)=1520-4x^2. Find the price (x)

Answer 1

f(x) = R(x)= 1520-4x^2 with x=price, would mean set x=0 and you maximize revenue at 1520, but that makes no sense. Where does the revenue come from if your sales are zero?

x (or q for quantity) usually equals output level, not price in dollars, then price = R/x. that doesn't work either, as the following shows. So, odds are you left out an x term. (see the end paragraphs of this answer)

Revenue = 1520-4x^2 with x=output, number of bicycles sold

Price times x = revenue

Px = R

P =R/x = 1520/x -4x

price that maximizes revenue is found by taking the derivative and setting it equal to zero

P'(x) = -1520/x^2 -4 = 0

-1520 -4x^2 = 0

4x^2 =-1520

x^2 =-1520/4 = -380

x= sqr(-380) = isqr380 = about 19 1/2 imaginary number of bicycles

the problem with the given quadratic equation you have is that with zero bicycles sold, revenue = 1520 and each bicycle sold reduces revenues.

Revenue maximizing output level is zero, unless you sell "imaginary" numbers of bicycles.

My guess is you left out the x term and mixed up output with price. Revenue probably was 1520x -2x^2 where x=output

R= 1520x -2x^2 then

R'= 1520-4x=0

4x=1520

x= 380 bicycles maximizes revenue

Price = R/x = 1520-2x = 1520-2(380) = 1520-760= 760

Price = $760 = revenue maximizing price for the bicycle.