I’ve been sick over the past two weeks with a chest infection, so I’ve had a mixture of no thoughts and a bunch of random and seemingly non sequitur practical applications of theoretical models. It made me consider running a series of optimisation posts and how they can be applicable to modern day businesses, processes, and selling techniques.

This series will look at various optimisation processes, including matrices, regression analysis, and quadratic equations. And this is where we’ll be starting off – quadratic equations. I guess this concept has been running around in the back of my head as I have been discussing offering mathematics and humanities private tuition to people in my local area, alongside private music tuition. Seeing that many of the young people will stay in the local community after graduating from high school (in grade 10 in Tasmania), I thought it was a nice opportunity to combine this post for my personal work and how it could assist those locally.

Quadratic Equations

Businesses can use demand curves to estimate how much potential profit can be made from certain products. Given that mountain bike riding is a very popular sport where I currently live, hosting some world class tracks, I’ve decided to choose this industry to gain insights into some of the local oligopolist businesses.

If one of them decided to manufacture a new bike, what kind of profit might they be looking to receive, given a particular selling price?

The costs have been estimated to be:

• $700,000 (advertising, manufacturing, etc), and
• $110 (cost of goods sold; cost to make a bike).

Given what they sell in stores, they can expect sales to follow a similar demand curve, so we can estimate this curve to be:

Sales = 70,000 − 200P,

where P is the price of a bike.

To showcase the use of this curve at the following prices, they can expect to sell the number of bikes:

• $0, they will give away 70,000 bikes (Sales = 70,000 – 200*0 = 70,000)
• $350, they will sell 0 bikes (Sales = 70,000 – 200*350 = 70,000 – 70,000 = 0)
• $300, they will sell 10,000 bikes (Sales = 70,000 − 200*300 = 70,000 – 60,000 = 10,000)

Given that we have this range of bike and price information, what is the best bike price the business should set to maximise their profits?

How many you sell depends on price, so use “P” for Price as the variable

Profit = Sales – Costs

Unit Sales

= 70,000 − 200P

Sales in Dollars

= Units × Price

= (70,000 − 200P) × P

= 70,000P − 200P²

Costs

= 700,000 + 110 x (70,000 − 200P)

= 700,000 + 7,700,000 − 22,000P

= 8,400,000 − 22,000P

Profit

Sales – Costs

= 70,000P − 200P² − (8,400,000 − 22,000P)

= −200P² + 92,000P − 8,400,000

Now that we have the profit curve, it can tell us 3 interesting figures:

1. 2 prices where the number of bike sold yield $0 profit, and
2. The number of bikes that maximise profit.

To best work with this function, I will divide everything by -200:

Profit = −200P² + 92,000P − 8,400,000

Profit = P² – 460P + 42,000

1. Solving for the 2 prices that yield $0 profit:

Solving for the roots, we can enter all information into the quadratic formula:

(-b -/+ sqrt(b^2 – 4*a*c))/2a

(-(-460) -/+ sqrt((-460)^2 – 4*1*(42,000)))/2*1

(460 -/+ sqrt(211,600 – 168,000))/2

(460 -/+ sqrt(43,600))/2

(460 -/+ 208.806)/2

x₁ = ($460 – $208.806)/2 = $125.597

x₂ = ($460 + $208.806)/2 = $334.403

So, if the bike price is set to either $125.597 and below or $334.403 and above, the business will generate zero to negative profit.

So, what’s the price that maximises profit?

2. Price to maximise profit:

Profit = P² – $460P + $42,000

In order to maximise profit, it is possible to take the derivative of the profit curve, set the result to zero and solve for P:

Profit = P² – $460P + $42,000

Profit’ = 2P – $460

0 = 2P – $460

P = $460/2

P = $230

To check if this is the correct value, we know that we can find the middle number of the profit curve, using the numbers found in 1.

Average of the range of x₁ and x₂ added to the value x₁:

(($334.403-$125.597)/2)+$125.597 = $230.

Profit

Sales – Costs

= $70,000*$230 − $200*($230)² − ($8,400,000 − $22,000*($230))

= −$200*($230)² + $92,000*($230) − $8,400,000

= -$10,580,000 + $21,160,000 − $8,400,000

= $2,180,000

So, it would be my professional opinion, based on previous data and insights into the local recreation sports sector, that the business price their newly manufactured bikes at $230 to maximise their profits for $2,180,000.