The demand equation for a certain product is given by p 138

The demand equation for a certain product is given by p = 138 0.09 x p=138-0.09x , where p p is the unit price (in dollars) of the product and x x is the number of units produced. The total revenue obtained by producing and selling x x units is given by R = x p R=xp. Determine prices p p that would yield a revenue of $7610. Note: you are asked for prices, not quantities. When you do get your quantities, you can round to the nearest item to get your prices.

Lowest such price:

Highest such price:

Solution

Let x be the number of units required to be produced and sold for generating a revenue of $ 7610. Then 7610 = xp or x = 7610/p. On substituting this value of x in the demand equation, we get p = 138 - 0.09(7610/p ). Now, on multiplying both the sides by 10p, we get 10p²= 1380p - 6849 or, 10p² - 1380p + 6849 = 0. Then, by the quadratic formula, we have p = [ -(-1380) ± { ( -1380)² -4*10* 6849} ]/ 2*10 = [ 1380± ( 1904400 - 273960) ]/ 20 = [ 1380 ± 1630440]/20 = (1380± 1276.890/20. Thus, either p = (1380 + 1276.89 / 20 = 2656.89/20 = $ 132.84 OR p = ( 1380 - 1276.89)/20 = 100.11/20 = $ 5.01 Thus, the lowest such price is $ 5.01 and the highest such price is $ 132.84..