# How To Write A Formula For An Exponential Function Calculus Applications in Real Estate Development

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## Calculus Applications in Real Estate Development

Calculus has many real-world applications and applications in the natural 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 applying mathematical models to speculative real estate development (ie: new house construction). Logically, a new home builder wants to make a profit after the completion of each home in the new home area. The builder will also need to be able to maintain (hopefully) good cash flow during each home building process, or each stage of home development. There are many factors that go into calculating profit. For example, we already know how to make a profit by: P = R – Cthat is, the profit (P) is equal to the income (R) subtract value (C). Although this formula is very simple, there are many variables that can affect this formula. For example, under cost (C), there are many different costs, such as the cost of construction materials, labor costs, maintenance 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 that must be considered in the formula mentioned above. Under income (R), one can include variables such as a lower home sale price, additional improvements or additions to the home (security system, surround sound system, granite countertops, etc.). Just plugging in all these different variables in and of itself can be a daunting task. However, this becomes more difficult if the rate of change is not linear, we need 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 another place where calculus comes into play.

Let’s say, last month we sold 50 homes for an average price of \$500,000. Excluding other factors, our revenue (R) is the price (\$500,000) times x (50 homes sold) equals \$25,000,000. Let us consider that the total cost of building 50 houses was 23,500,000 dollars; 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 increase profits for the next month. How do you do this? What price would you put on it?

As a simple example of this, let’s first calculate the marginal gain in terms of x to build a home in a new residential area. We know that income (R) equals the demand equation (ptimes units sold (x). We write the equation as

R = px.

Let’s say we determine the demand equation for selling a home in this area

p = \$1,000,000 – x/10.

With \$1,000,000 you know you won’t sell any houses. Now, the value equation (C) is him

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

From this we can calculate the minimum profit in terms of x (units sold), then use marginal profit to calculate the price we should charge to maximize profit. So, income i

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

So, profit

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 gain by taking the derivative of the gain

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 find the following:

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

So, the price we need to set to get the maximum profit for each house we sell should be \$509,000. The next month you sell 50 more homes with a new price structure, and you earn an increase in revenue of \$450,000 from the previous month. Great job!

Now, next month your boss is asking you, a community builder, to find a way to reduce the cost of building a home. Since before you know that the equal value (C) it was:

\$300,000 + \$18,000x (\$175,000 in fixed material 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 cost of fixed materials to \$150,000 and \$9,000 per house, and reduce your operating costs to \$110,000 and \$7,000 per house. Hence the balance of your costs (C) changed to

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

Because of this change, you will need to recalculate the profit base

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 gain by taking the derivative of the new calculated gain

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 find the following:

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

So, the price we need to set to get the maximum profit for each house we sell should be \$508,000. Now, although we reduce the selling price from \$509,000 to \$508,000, and we still sell 50 units like two months ago, our profit increases because we reduce the cost to \$140,000. We can find this by calculating the difference between the first P = R – C and the second P = R – C which contains the equivalent of new costs.

1 st 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

Taking the second profit minus the first profit, you can see a difference (increase) of \$140,000 in profit. So, by reducing the cost of building a home, you are able to make the company more profitable.

Let’s refresh again. Just by using the demand function, marginal profit, and maximum profit from the calculus, and nothing else, you were able to help your company increase the monthly income of the ABC Home Community project by hundreds of thousands of dollars. With a little negotiation with your building suppliers and labor leaders, you can lower your costs, with a simple adjustment of the cost equation (C), you can quickly see that by reducing costs, you have increased profits again, even after adjusting your gross profit by reducing 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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