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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 on 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 (eg: new home construction). Logically, a new home builder wants to make a profit after the completion of each home in a new home community. This builder will also (hopefully) need to be able to maintain a positive cash flow during the building process of each home, or each phase of home development. There are many factors involved in calculating a profit. For example, we already know that the formula for profit is: P = R – Cthat is, the profit (P) is equal to the income (R) minus the cost (C). Although this primary formula is very simple, there are many variables that can factor into this formula. For example, under cost (C), there are many different variables of cost, such as the cost of building materials, cost of labor, holding cost of real estate before purchase, utility cost and insurance premium cost during the construction phase. These are some of the many costs to factor into the above formula. Under income (R), one can include variables such as the base sales price of the home, additional upgrades or additions to the home (security system, surround sound system, granite countertops, etc.). Just plugging in all these different variables in itself can be a daunting task. However, it 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 form of a curve (ie: exponential rate of change)? This is one area where calculus comes into play.

Let’s say we sold 50 homes last month with an average sale price of $500,000. Excluding other factors, our income (R) is price ($500,000) times x (50 homes sold) which equals $25,000,000. Let’s consider that the total cost to build all 50 houses was $23,500,000; so the profit (P) is 25,000,000 – $23,500,000 which equals $1,500,000. Now, knowing these numbers, your boss has asked you to maximize the profit for the next month. How do you do it? 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 house in a new residential community. We know that income (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 is for the sale of a home in this community

p = $1,000,000 – x/10.

At $1,000,000 you know you won’t sell any houses. Now, the cost equation (C) is

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

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 income is

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

Therefore the profit

P = R – C = ($1,000,000x – x^2/10) – ($300,000 + $18,000x) = 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 to the request function and get the following:

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

So, the price we need to set to get the maximum profit for each home we sell must be $509,000. The next month, you sell 50 more homes with the new pricing structure, and get a $450,000 profit increase from the previous month. Good work!

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

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

After clever negotiations with your building suppliers, you were able to reduce the fixed material costs to $150,000 and $9,000 per home and lower your labor costs to $110,000 and $7,000 per home. As a result, your cost equation (C) changed to

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

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

P = R – C = ($1,000,000x – x^2/10) – ($260,000 + $16,000x) = 984,000x – (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 to the request function and get the following:

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

So, the price we need to set to get the new maximum profit for each house we sell must 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 still increased because we cut costs to $140,000. We can figure 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 = ($1,000,000x – x^2/10) – ($300,000 + $18,000x) = 982,000x – (x^2/10) – $300,000 = 48,799,750

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

If you take the second profit minus the first profit, you can see a difference (increase) of $140,000 in profit. So, by cutting home building costs, you can make the company even more profitable.

Let’s resume. 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. Through a little negotiation with your building suppliers and labor leaders, you were able to lower your costs, and through a simple readjustment of the cost equation (C), you could quickly see that by cutting costs, you increased profits 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.

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