# How Long Does It Take To Adjust To New Formula 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 the physical sciences, computer science, economics, business, and medicine. I will briefly discuss some of these uses and applications in the real estate industry.

To begin with, we will use the calculation examples of speculative real estate development (i.e. new housing construction). It makes sense that a new home builder wants to make a profit after each new home community home is completed. Also, this builder must be able to maintain (hopefully) positive cash flow at each stage of the home construction process or home development. There are many factors that go into calculating profit. For example, we already know that the profit formula is: P = R – Cwhat is the profit (p) is equal to the income (R) minus cost (C). Although this basic formula is very simple, there are many variables to this formula. For example below cost (C), there are many different cost variables, such as the cost of construction materials, labor costs, property holding costs before purchase, utility costs and insurance premium costs during the construction phase. These are some of the many costs to factor into the above formula. Below income (R), can include variables such as the home’s base sales price, additional home upgrades or add-ons (security system, surround sound system, granite countertops, etc.). Connecting all these different variables in itself can be a daunting task. But it gets even 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 the variables is curve-shaped (ie, an 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 revenue (R) is price (\$500,000) times x (50 homes sold) equals \$25,000,000. We estimate that the total cost to build all 50 homes was \$23,500,000; so profit (p) is \$25,000,000 to \$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 that? What price can you set?

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

R = px.

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

page = \$1,000,000 – x/10.

For \$1,000,000 you know you won’t be selling any homes. Now the cost equation (C) is

300,000 + \$18,000x (\$175,000 fixed materials cost 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. So there is income

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

So the profit is

p = RC = (\$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 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 connect x back to the demand function and get the following:

page = \$1,000,000 – (4,910,000)/10 = \$509,000.

So the price for each house sold should be set at \$509,000 to get the maximum profit. Next month, you sell 50 more homes with the new pricing structure, and the net profit increases by \$450,000 over the previous month. Excellent job!

Next month, your boss asks you, the community developer, to find a way to reduce the cost of home construction. Before you knew that the cost equation (C) was:

300,000 + \$18,000x (\$175,000 fixed materials cost 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 materials costs to \$150,000 and \$9,000 per house and labor costs to \$110,000 and \$7,000 per house. As a result, your cost equation (C) has become the following

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

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

p = RC = (\$1,000,000xx^2/10) – (\$260,000 + \$16,000x) = 984,000x – (x^2/10) – \$260,000.

Based on 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 connect x back to the demand function and get the following:

page = \$1,000,000 – (4,920,000)/10 = \$508,000.

So each house sold would need to be priced at \$508,000 to generate a new maximum profit. Now, even though we reduce the selling price from \$509,000 to \$508,000 and still sell 50 units as in the previous two months, our profit is still increased because we cut costs to \$140,000. We can find this out by calculating the difference between the first P = R – C and another P = R – C which includes a new cost equation.

1 p = RC = (\$1,000,000xx^2/10) – (\$300,000 + \$18,000x) = 982,000x – (x^2/10) – \$300,000 = 48,799,750

2 p = RC = (\$1,000,000xx^2/10) – (\$260,000 + \$16,000x) = 984,000x – (x^2/10) – \$260,000 = 48,939,750

By taking the second profit minus the first profit, you see a profit difference (increase) of \$140,000. Thus, by reducing the cost of home construction, you can make the business even more profitable.

Let’s summarize. By simply applying the demand function, marginal profit, and maximum profit calculation and nothing else, you helped your company increase its monthly profit from the ABC Home Community project by hundreds of thousands of dollars. With a little negotiation with your construction suppliers and construction managers, you were able to lower your costs and by simply readjusting the cost equation (C), you could quickly see that by cutting costs, you increased profits again, even after adjusting for the maximum profit, lowering your selling price by \$1,000 per unit. This is an example of the wonder of computation in solving real-world problems.

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