# All Of The Above Are Flow Variables Previous Next Calculus Applications in Real Estate Development

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## Calculus Applications in Real Estate Development

Calculus has many real-world uses and applications in physics, 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 (eg: new home construction). Logically, a new home builder wants to make a profit after each home is completed in a new home community. This builder must be able to maintain a positive cash flow during each home construction process or each stage of home development. There are many factors to measure profitability. For example, we already know that the profit formula is: P = R – Ci.e. profit (P) is equal to the revenue (R) minus expenses (C). Although this basic formula is very simple, there are many variables involved in this formula. For example, at low cost (C), there are many different cost variables, such as the cost of construction materials, labor costs, holding costs of real estate before purchase, utility costs, and insurance premium costs during the construction phase. These are some of the many costs to consider in the formula mentioned above. Under Earnings (R), can include variables such as the home’s original sales price, additional upgrades or add-ons (security systems, surround sound systems, granite countertops, etc.) to the home. Just plugging in all these different variables in and of itself can be a daunting task. However, this becomes more 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 curvilinear in shape (ie: 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 sale price of \$500,000. Without considering other factors, our revenue (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; Hence 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 increase the profit for the next month. How do you do this? What price can you set?

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

R = px.

Suppose we have determined that there is a demand equation for selling houses 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 x (units sold), then use marginal profit to calculate what price we should charge to maximize profit. So, there is the revenue

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

It makes profit

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

From this we can derive the marginal profit by taking the profit derivative

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 Return to the demand function and get the following:

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

Therefore, the price we set should be \$509,000 to maximize profit for each home we sell. The following month you sell 50 more homes with the new pricing structure and increase net profit by \$450,000 over the previous month. Good work!

Now, for the next month your boss asks you, the community developer, to find a way to reduce the cost of building a house. You already 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 shrewd negotiations with your building suppliers, you were able to reduce fixed materials costs to between \$150,000 and \$9,000 per home, and reduce your labor costs to between \$110,000 and \$7,000 per home. Consequently your cost equation (C) is changed to

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

Due to these changes, you will have to recalculate the original profit

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

From this we can extract new marginal profits.

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 Return to the demand function and get the following:

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

Therefore, the price we set to achieve the new maximum profit for each home we sell should be \$508,000. Now, even if we reduce the selling price from \$509,000 to \$508,000 and we still sell 50 units as in the previous two months, our profit still increases because we cut costs by \$140,000. We can find this by calculating the difference between the first P = R – C and another P = R – C which contains the new cost equation.

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

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

If you subtract the first profit and take the second profit, you can see a difference (increase) in profit of \$140,000. So, by reducing the cost of house construction, you can make the company more profitable.

Let’s summarize. Using only the demand function, marginal profit and profit maximization calculus, and nothing else, you can help your company increase its monthly profits from the ABC Home Community Project by hundreds of thousands of dollars. With a little negotiation with your building supplier and labor leaders, you were able to lower your costs and with a simple modification of the cost equation (C), even after adjusting your maximum profit by reducing your selling price by \$1,000 per unit, you have again increased profit by cutting costs. This is an example of the wonder of calculus when applied to real-world problems.

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