Suppose on a highway, there is a Dhaba. Name it by Dhaba A.

You are also planning to set up a new Dhaba. Where will you set up your Dhaba?

Model this as a Mathematical Problem. This is an interesting and creative part of the BusinessoMath-man in you.

You have to assume something for Mathematical simplicity to model a real life phenomenon via math.

Assumptions:

  • The length of the highway is 1 unit. Assume the highway is denoted by the interval [0,1].
  • The number of Persons in a road is proportional to the road length. P = c. R, where P is the Persons and R is the road length, c is an arbitrary constant.
  • The profit is modeled as the Total Number of Persons arriving in the Dhaba.
  • A person will visit a dhaba which is nearer over another which is further away.
  • When a customer is at the same distance from two shops then the customer choose randomly.
  • Your aim is to maximize your profits.

Observe that the assumptions are valid and are actually followed in real life. Just sit and think for some time placing yourself in that position.

The Explicit Mathematical Problem

The Highway Modelled as [0,1]
The Dhabba A is placed at a distance d say from origin. The distance d is fixed.
Suppose you place Dhaba B at a distance x from the origin. The distance x is variable and our aim is to find an appropriate x. Assume 0=<x<=d

Profit Calculation of your Dhaba B

Now based on the lengths of the roads above and the assumptions made, we have to calculate the profit of Dhaba B.

Observe how the people and customers behave according to the positions of theirs and according to the assumptions.

So the profit made by B = c.R , where R is the length of the road on which B has monopoly. Here, as shown R = \c.( x + \frac{|d-x|}{2}\).

We have assumed in this case that \(0 \leq x \leq d\) by the diagram.

So, the profit for the Dhaba B is c.\(\frac{d+x}{2}\) .

Hence as \(0 \leq x \leq d\), the profit is maximized if x = d.

Exercise: Show that the profit of Dhaba B as a function of x is c.(\( 1 – \frac{d+x}{2}\)) if \(d \leq x \leq 1\) .

Also, observe that in this case the profit is maximized at x = d.

Exercise: Calculate the maximum profit in both cases. Do you observe something fishy? What steps and what arguments will you give to understand the fishiness?

Please share your views in the comments section.

Eager to listen to your beautiful ideas.