How Calculus Can Help You in Business

Ever wonder why we take calculus during college (with the exception of degrees heavily focusing in the arts)? I remember that most students do not really appreciate this subject due to the 'alien' concept and mind-boggling equations.

Being an engineering student myself, I was a bit confused as to how this subject can help me once I get a full time job. Like any other student, I also asked sarcastically the question - Why do I need calculus? It's not like I'll go to the market and ask for the differential price of a product. Nobody really uses it in real life.  

But as years gone by, and I matured in my chosen field of profession - I realised that Calculus was more than a math subject I need to pass in college. I remember my CEO calling me one time showing me a bell curve graph showing the fluctuation of our EBITDA numbers as we increase or decrease our OPEX. I know, I know... big words! Simply put, my CEO was asking me to find out what metric should we tweak in order to meet that sweet spot to optimise our production cost and at the same time hit our profitability target.

This experience is one of the countless events that forced me to go back to my old textbooks and find a way to answer these crucial questions. And it brought me back to Calculus and the concept of derivatives, limits and functions.

In order to appreciate Calculus and its many applications for business, let me hone this discussion in the topic of derivatives and differentiation. To make it simpler, let me provide a non-standard definition of these two concepts in the context of business administration and management.

A DERIVATIVE is something (a number, a metric, a KPI, etc) which is based  on another source. It is the number or the event that you would like to know given a certain set of information or facts to predict a certain outcome. It is determined only if another variable is present. And the process of finding this Derivative is called DIFFERENTIATION. 

To make sense of this topic, let me give you two cases that I've used  this concepts in reality

CASE 1: Budget Rooms for Rent

Situation: One of the business owners I've helped is an owner of a small apartment for rent for backpackers. He owns 250 rooms across the city and marketing it heavily via the internet and travel agencies. If they rent X apartments then his monthly profit, in dollars, is given by the following equation:

P(x) = 3200x - 80,000 - 8x² 

Where:
3,200x is the price per room per stay.
-80,000 is the fixed monthly cost to operate the business.
- 8x² is the recorded rate of the variable cost during operations.

Case Problem: How many rooms should they rent in order to maximise profit?

Some people would think immediately that renting all rooms will be the best case scenario reaping maximum profits. But factoring in cost of operations (both fixed and variable cost), it would be interesting to know if renting out all rooms will definitely be the best possible course monthly for this business owner.

Solution:

We know that the maximum profit is between 0<x<250 range of rented rooms. What we need to do next is to find the derivative of the profit as declared above to get the Critical Points to which we will get the maximum profit. Again, as defined, derivative is something that you would like to know based on other facts (in this case the PROFIT). And we need to differentiate it to get the derivative.

(I assume you know basic differentiation)

P'(x) = 3200-16X
If P'(x) = 0
0 = 3200-16X
16X=3200
X = 200

200 is one of the critical points. Hence the range is between 0<200<250

Substituting these three numbers, we get the following profit per month:

P(0) = -80,000
P(200) = 240,000
P(250) = 220,000

As you can see, renting all 250 rooms produced less profits versus 200 rooms if we factor in operational cost. Hence, the business owner was advice to aim to rent out 200 rooms at regular price in any given month and utilise the remaining 50 for any promotions they would like to do to further increase their earnings.

CASE 2: Cupcakes for Sale!

Situation: Every December, one of my friends sells cupcakes to earn extra during the holiday season. One of her biggest challenge is that after the end of the month, she feels that her total effort in producing the cupcakes versus the money she earns from it is not generating her significant income. She doesn't know what price tiering she would do because her cost increases if she produces more cupcakes.

So, after observing and recording data, we found out that her production cost per day has this function:

C(x) = 2500 - 10x - 0.01x²  - 0.0002x³
Where:
2500 is her total cost
-10X is her savings since she can produce 10 more cupcakes using the same amount of ingredients
- 0.01x² is her savings on electricity
- 0.0002x³ is her savings on delivery per batch to clients

Case Problem: She wanted to know what would be the rate of change of her cost if she produces 200, 300 and 400 cupcakes in a day in order for her to create an optimal pricing scheme.

Solution:

Same process in the first case, look for the derivative and apply the critical points 200, 300 and 400.

C'(x) = -10 - 0.02x + 0.0006x² 

C'(200) = $10
C'(300) = $38
C'(400) = $78

As you can see, producing the 201st cupcake will add $10 to her cost, 301st cupcake will cost $38 and the 401st cupcake will be $78.