4.3 Example using the retail space problem

Next: 4.4 Up: 4. Previous: 4.2

4.3 Example using the retail space problem

Carry on with the retail example, but this time ask the question: Find the retail shelf space of the new product needed to maximize additional sales.

Notice that this is a different question; remember that there is an alternative product available that guarantees $50 for every foot of shelf space.

Table 2: Relationship between sales of the new product, sales lost due to not using the alternative product, total additional sales, and the number of feet dedicated to its display


Feet	Sales of	Sales lost due to	Total
	new product	alternative product	additional sales
1	87.88	50	37.88
2	171.18	100	71.18
3	237.50	150	87.50
4	286.84	200	86.84
5	319.20	250	69.20
6	334.58	300	34.58
7	332.98	350	-17.02
8	314.40	400	-85.60
9	278.84	450	-171.16
10	226.30	500	-273.70

Clearly the answer is somewhere between 2 and 4 feet. Find the answer using the calculus. Notice that the total additional sales (call that T) equals

$\begin{displaymath}\fbox{$T = \overbrace{-12.40 + 108.77\,L - 8.49\,L^2}^{\mbox{... ...es}} \quad - \quad \overbrace{50\,L}^{\mbox{\rm Lost sales}}$ }\end{displaymath}$

Find the function you want to optimize.

$\begin{displaymath}\fbox{$T = -12.40 + 108.77\,L - 8.49\,L^2 - 50\,L$ }\end{displaymath}$

Differentiate it.

$\begin{displaymath}\fbox{$\frac{dT}{dL} = 108.77 - 2 \times 8.49\,L - 50$ }\end{displaymath}$

Set the derivative equal to zero.

$\begin{displaymath}\fbox{$0 = 108.77 - 2 \times 8.49\,L - 50$ }\end{displaymath}$

Solve for the quantity of interest.

$\begin{displaymath}\fbox{$L = 3.461$ }\end{displaymath}$

So you need to display about three and a half feet in order to maximize total additional sales.

Next: 4.4 Up: 4. Previous: 4.2

Richard Waterman
1999-06-14