Mary Jane, the confectioner with two first names, looked up her fellow candy maker in the phone book. “Cracker, Jack” was the listing, and during the ensuing phone conversation they decided to market their products jointly in a single multi-pack. They quickly discovered that the normal rules of economics applied, as when they reduced their price, they were able to sell more of their product, and vice versa.
When they priced a box of candy at $5.95, they realized a profit of $1.75 per box and sold on average 100 boxes per week.
For each $0.20 reduction in price, their sales increased by 25 boxes per week.
What did they finally determine was the optimum price at which to sell their wares to produce the maximum weekly profit?
(Assume that changes in price have no effect on the cost of producing or marketing the candy.)
Answer
This can be solved in a few different ways. Trial and error is one, and perhaps the simplest. Deriving an equation for the profit produces:
P(Profit) = ($1.75-0.20x)(100+25x)
…….. = -5x²-20x+43.75x+175
…….. = -5x² +23.75x +175
Here, the ‘x’ represents the number of $0.20 reductions in price (and per-box profit), and thereby the number of 25 box sales increases that will be performed.
This produces an inverted parabola when graphed, and the point of maximum profit can be fairly accurately determined by observation. Finally, using basic calculus, the first derivative can be obtained:
P` = -10x +23.75
To maximize profit, find the point where the first derivative is equal to zero.
0 = -10x +23.75
10x = 23.75
x = 2.375
Therefore, the price should be reduced by 2.375 x $0.20, or $0.47 per box for an optimum price of $5.48 per box.
This yields sales of approximately 159 boxes per week, at a profit per box of $1.28 for a total weekly profit of $203.52.
