New Port Beach Business Daily

How do I solve this revenue problem by completing the square?

A sportswear store sells baseball caps with the local baseball team's logo on them. Last year the store sold 600 caps at $15 each. The store manager is planning to increase the price. A consumer survey shows that for every $1 increase there will be a drop of 30 sales a year. a) What should the selling price be to maximize the annual revenue? b) What is the maximum revenue from the caps? How do I figure it out? Please help!

Public Comments

1. Hello Julia, I'll assign some variables: Let R be revenue per year, in dollars. Then R = units sold * price per unit. Let units sold be U caps. Let price per unit be P, in dollars per cap. Thus: R = UP. Note that to begin with, P = $15 per cap, and U = 600 caps. To increase the price by $1, and drop sales by 30: R = (600 - 30) (15 + 1). To do the same thing again: R = (600 - 30 - 30) (15 + 1 + 1) = (600 - 60) * (15 + 2). Notice that these will decrease by multiples of 30 and increase by multiples of 1 respectively. Let's give the number of repetitions the value n. So: U = 600 - 30n. P = 15 + n. So: R = UP = (600 - 30n) (15 + n). Expanding: R = 9000 + 600n - 450n - 30n². R = 9000 + 150n - 30n². Factorising: R = -30 [n² - 5n - 30 ]. Completing the square: R = -30 [ (n - 2½)² - (2½²) - 30 ]. R = -30 [ (n - 2½)² - 36¼ ]. R = 30*36¼ - 30(n - 2½)². [ NOTE: If you have not been taught differentiation, do not panic. Since you have completed the square, any derivative will only be 0 at the point when the square is 0. i.e. (n - 2½)² must equal 0. (n - 2½)² = 0. Square rooting: n - 2½ = 0. n = 2½. I have included the differentiation below to assert the theorem.] Differentiating: dR/dn = -60(n - 2½) = 150 - 60n. Solving for 0: dR/dn = 0 → 150 - 60n = 0. 150 = 60n. n = 15/6 = 2½. So, for (a) : Let n = 2½. P = 15 + n = 15 + 2½ = 17½. Thus P = $17.50 per cap. For (b) : Again, n = 2½. R = (600 - 30n) (17½). R = (600 - 30(2½)) (17½). R = (600 - 75) (17½). R = 525 * 17½. R = $9187.50 per year.
2. Let the new price be x then loss of sales=(x-15)*30 loss of revenue=(x-15)*30*x Net revenue=R=600x-((x-15)30)x = 600x-30x^2+450x At the point of maximum revenue dR/dx ie the first derivative will be zero: dR/dx=60x+600+450=0 which gives x=17.5 Hence maximum revenue will be if price is increased to $17.5 Any increase beyond $17.5 or decrease below $17.5 will give a lower net revenue. maxm. revenue can be $9187.50