# Enter The Formula That Will Calculate The Total Projected Revenue 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 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 (ie: new home construction). Logically, a new home builder wants to make a profit after completing each home in a new home community. This builder needs to be able to (hopefully) maintain a positive cash flow during each home construction process or each stage of home development. There are many factors that go into calculating profits. For example, we already know the profit formula is: P = R – Cie profit (P) revenue is equal to (R) minus cost (c). Although this basic formula is very simple, there are many variables that can be factored into this formula. For example, under cost (c), there are many cost variables, such as construction material costs, labor costs, holding costs of real estate before purchase, utility costs, and insurance premium costs during the construction phase. These are just a few of the many costs to factor into the above formula. Under Revenue (R), one can include variables such as the home’s base sales price, additional upgrades or add-ons to the home (security systems, surround sound systems, granite countertops, etc.). Plugging in all these different variables can be a daunting task in itself. However, it gets more complicated if the rate of change is not linear, so we need to adjust our calculations because the rate of change of one or all of these variables is in the shape of a curve (eg: 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 sales price of \$500,000. Excluding other factors, our revenue (R) is the 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 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 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 x Building a home in a new residential community. We know that revenue (R) the demand equation equals (p) times units sold (x). We write the equation as

R = px.

Suppose we have determined the demand equation for the sale of 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 fixed materials costs and \$10,000 per house sold + \$125,000 fixed labor costs and \$8,000 per house).

From this we can calculate the marginal profit x (units sold), then use marginal profit to calculate the price we should charge to maximize profit. Therefore, there is revenue

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

So, profit is

P = R – C = (\$1,000,000xx^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 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 Return to the demand function and get the following:

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

So, the price we need to set to make the maximum profit for each home we sell should be \$509,000. The next month you sell 50 more homes with the new pricing structure, and a net profit increase of \$450,000 over the previous month. Good job!

Now, for the next month your boss asks you to find a way to cut the cost of building a house for the community developer. You know that the cost equation (c) was:

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

After shrewd negotiations with your construction suppliers, you were able to reduce fixed material costs between \$150,000 and \$9,000 per home, and your labor costs between \$110,000 and \$7,000 per home. As a result your cost equation (c) has changed to

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

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

P = R – C = (\$1,000,000xx^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 calculated new profit.

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.

So, the price we need to set to earn the new maximum profit for each home we sell must be \$508,000. Now, even though we reduced the selling price from \$509,000 to \$508,000, and we still sold the same 50 units as the previous two months, our profit still increased because we reduced costs to the tune of \$140,000. We can find this by first calculating the difference between P = R – C And secondly P = R – C which includes a new cost equation.

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

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

By subtracting the first profit and taking the second profit, you can see a difference (increase) in profit of \$140,000. So, by cutting costs in home construction, you are able to make the company more profitable.

Let’s summarize. By simply applying the demand function, marginal profit, and profit maximization calculus, and nothing else, you were able to help your company increase its monthly profits from the ABC Home community project by hundreds of thousands of dollars. By a little negotiation with your manufacturing suppliers and labor leaders, you were able to lower your costs, and by a simple readjustment of the cost equation (c), you can immediately see that by cutting costs, you still increased profits, even after adjusting your maximum profit by reducing your selling price by \$1,000 per unit. This is an example of the wonders of calculus when applied to real-world problems.

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