He predicts that he will sell 280 bicycles for $ 360 . He also predicted that for each increase in the price of $ 10, he will sell 5 bikes less.

Question

How many bicycles will he sell in this context?

what is the maximum price that he can obtain?

what will be the maximum revenue?

Answer 1

Let x = number of $10 increases

 

R(x) = revenue = (price)(quantity)

 

                      = (360 + 10x)(280 - 5x)

 

The graph of R(x) is a parabola opening downward.  The x-intercepts occur when 360 + 10x = 0 and when 280 - 5x = 0.  So, the x-intercepts are (-36,0) and (56,0).

 

By the symmetry of the parabola, the maximum revenue occurs when x is halfway between the x-intercepts.  So, we have maximum revenue when x = (-36+56)/2 = 20/2 = 10.

 

To maximize revenue, 10 ten dollar increases are required.  So, the price per bicycle should be $460.  At that price, the number of bicycles sold is predicted to be 230.  The maximum revenue would then be       ($460/bicycle))(230 bicycles) = $105,800.