# Calculus Applications in Real Estate Development

Calculus has many real world uses and applications in the physical sciences, computer science, economics, business, and medicine. I will briefly touch upon some of these uses and applications in the real estate industry.

Let’s start by using some examples of calculus in speculative real estate development (i.e.: new home construction). Logically, a new home builder wants to turn a profit after the completion of each home in a new home community. This builder will also need to be able to maintain (hopefully) a positive cash flow during the construction process of each home, or each phase of home development. There are many factors that go into calculating a profit. For example, we already know the formula for profit is: *P = R – C*, which is, the profit (*P*) is equal to the revenue (*R*) minus the cost (*C*). Although this primary formula is very simple, there are many variables that can factor in to this formula. For example, under cost (*C*), there are many different variables of cost, such as the cost of building materials, costs of labor, holding costs of real estate before purchase, utility costs, and insurance premium costs during the construction phase. These are a few of the many costs to factor in to the above mentioned formula. Under revenue (*R*), one could include variables such as the base selling price of the home, additional upgrades or add-ons to the home (security system, surround sound system, granite countertops, etc). Just plugging in all of these different variables in and of itself can be a daunting task. However, this becomes further complicated if the rate of change is not linear, requiring us to adjust our calculations because the rate of change of one or all of these variables is in the shape of a curve (i.e.: exponential rate of change)? This is one area where calculus comes into play.

Let’s say, last month we sold 50 homes with an average selling price of $500,000. Not taking other factors into consideration, our revenue (*R*) is price ($500,000) times x (50 homes sold) which equal $25,000,000. Let’s consider that the total cost to build all 50 homes was $23,500,000; therefore the profit (*P*) is 25,000,000 – $23,500,000 which equals $1,500,000. Now, knowing these figures, your boss has asked you to maximize profits for following month. How do you do this? What price can you set?

As a simple example of this, let’s first calculate the marginal profit in terms of *x* of building a home in a new residential community. We know that revenue (*R*) is equal to the demand equation (*p*) times the units sold (*x*). We write the equation as

*R = px*.

Suppose we have determined that the demand equation for selling a home in this community is

*p* = $1,000,000 – *x*/10.

At $1,000,000 you know you will not sell any homes. Now, the cost equation (*C*) is

$300,000 + $18,000*x* ($175,000 in fixed materials costs and $10,000 per house sold + $125,000 in fixed labor costs and $8,000 per house).

From this we can calculate the marginal profit in terms of *x* (units sold), then use the marginal profit to calculate the price we should charge to maximize profits. So, the revenue is

*R* = *px* = ($1,000,000 – *x*/10) * (*x*) = $1,000,000*x* – *x^2*/10.

Therefore, the profit is

*P* = *R – C* = ($1,000,000*x* – *x^2*/10) – ($300,000 + $18,000*x*) = 982,000x – (*x^2*/10) – $300,000.

From this we can calculate the marginal profit by taking the derivative of the profit

*dP/dx* = 982,000 – (*x*/5)

To calculate the maximum profit, we set the marginal profit equal to zero and solve

982,000 – (*x*/5) = 0

*x* = 4910000.

We plug *x* back into the demand function and get the following:

*p* = $1,000,000 – (4910000)/10 = $509,000.

So, the price we should set to gain the maximum profit for each house we sell should be $509,000. The following month you sell 50 more homes with the new pricing structure, and net a profit increase of $450,000 from the previous month. Great job!

Now, for the next month your boss asks you, the community developer, to find a way to cut costs on home construction. From before you know that the cost equation (*C*) was:

$300,000 + $18,000*x* ($175,000 in fixed materials costs and $10,000 per house sold + $125,000 in fixed labor costs and $8,000 per house).

After, shrewd negotiations with your building suppliers, you were able to reduce the fixed materials costs down to $150,000 and $9,000 per house, and lower your labor costs to $110,000 and $7,000 per house. As a result your cost equation (*C*) has changed to

*C* = $260,000 + $16,000*x*.

Because of these changes, you will need to recalculate the base profit

*P* = *R – C* = ($1,000,000*x* – *x^2*/10) – ($260,000 + $16,000*x*) = 984,000*x* – (*x^2*/10) – $260,000.

From this we can calculate the new marginal profit by taking the derivative of the new profit calculated

*dP/dx* = 984,000 – (*x*/5).

To calculate the maximum profit, we set the marginal profit equal to zero and solve

984,000 – (*x*/5) = 0

*x* = 4920000.

We plug *x* back into the demand function and get the following:

*p* = $1,000,000 – (4920000)/10 = $508,000.

So, the price we should set to gain the new maximum profit for each house we sell should be $508,000. Now, even though we lower the selling price from $509,000 to $508,000, and we still sell 50 units like the previous two months, our profit has still increased because we cut costs to the tune of $140,000. We can find this out by calculating the difference between the first *P = R – C* and the second *P = R – C* which contains the new cost equation.

1st *P* = *R – C* = ($1,000,000*x* – *x^2*/10) – ($300,000 + $18,000*x*) = 982,000*x* – (*x^2*/10) – $300,000 = 48,799,750

2nd *P* = *R – C* = ($1,000,000*x* – *x^2*/10) – ($260,000 + $16,000*x*) = 984,000*x* – (*x^2*/10) – $260,000 = 48,939,750

Taking the second profit minus the first profit, you can see a difference (increase) of $140,000 in profit. So, by cutting costs on home construction, you are able to make the company even more profitable.

Let’s recap. By simply applying the demand function, marginal profit, and maximum profit from calculus, and nothing else, you were able to help your company increase its monthly profit from the ABC Home Community project by hundreds of thousands of dollars. By a little negotiation with your building suppliers and labor leaders, you were able to lower your costs, and by a simple readjustment of the cost equation (*C*), you could quickly see that by cutting costs, you increased profits yet again, even after adjusting your maximum profit by lowering your selling price by $1,000 per unit. This is an example of the wonder of calculus when applied to real world problems.