# Calculation 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 refer to some of these uses and applications in the real estate industry.

Let’s start by using some calculation examples in speculative real estate development (i.e. building new homes). Logically, a new home builder wants to make a profit after the completion of each home in a new home community. This builder should also be able to (hopefully) maintain positive cash flow during the construction process of each home or each phase of the home development. There are many factors that go into calculating a profit. For example, we already know that the formula for profit is: P = R – C, which is the profit (P) equals income (R) minus the cost (VS). Although this main formula is very simple, there are many variables that can be included in this formula. For example, low cost (VS), there are many different cost variables, such as the cost of building materials, labor costs, pre-purchase real estate maintenance costs, utility costs, and premium costs. insurance during the construction phase. These are some of the many costs that must be factored into the formula mentioned above. Below income (R), variables such as the base sale price of the home, additional improvements or complements to the home (security system, surround sound system, granite countertops, etc.) could be included. Simply plugging all these different variables into yourself can be a daunting task. However, this is further complicated if the rate of change is not linear, which requires us to adjust our calculations because the rate of change of one or all of these variables is in the form of a curve (i.e .: exponential rate of change) . This is one area where calculation comes into play.

Let’s say last month we sold 50 homes with a median sales price of \$ 500,000. Regardless of other factors, our income (R) is the price (\$ 500,000) multiplied by x (50 houses sold) which equals \$ 25,000,000. Let us consider that the total cost to build the 50 houses was \$ 23,500,000; therefore the gain (P) is 25,000,000 – \$ 23,500,000, which is equal to \$ 1,500,000. Now, knowing these figures, your boss has asked you to maximize profits for the following month. How do you do this? What price can you set?

As a simple example of this, let’s first calculate the marginal profit in terms of X to build a home in a new residential community. We know that income (R) is equal to the demand equation (P) times units sold (X). We write the equation as

R = px.

Suppose we have determined that the demand equation for selling a house in this community is

P = \$ 1,000,000 – X/ 10.

At \$ 1,000,000 you know that you will not sell any houses. Now the cost equation (VS) it is

\$ 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 marginal profit in terms of X (units sold), then use marginal profit to calculate the price we must charge to maximize profit. So the income is

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

Therefore, the gain 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 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 go back to the demand function and get the following:

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

Therefore, the price we should set to get the maximum profit for each house we sell should be \$ 509,000. The next month, he sells 50 more houses with the new price structure and makes a net profit increase of \$ 450,000. compared to the previous month. Great job!

Now, over the next month, your boss asks you, the community developer, to find a way to cut home construction costs. Before knowing that the cost equation (VS) 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 astute negotiations with his construction suppliers, he was able to cut his fixed material costs to \$ 150,000 and \$ 9,000 per house, and lower his labor costs to \$ 110,000 and \$ 7,000 per house. As a result, your cost equation (VS) has changed to

VS = \$ 260,000 + \$ 16,000X.

Due to 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 benefit by taking the derivative of the new calculated benefit

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 go back to the demand function and get the following:

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

Therefore, the price that we should set to obtain the new maximum profit for each house we sell should be \$ 508,000. Now, even though we lowered the sale price from \$ 509,000 to \$ 508,000, and we continue to sell 50 units Like the previous two months, our profits have still increased because we cut costs to the tune of \$ 140,000. We can find out by calculating the difference between the first P = R – C and the second P = R – C A containing 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

Taking the second profit minus the first profit, you can see a difference (increase) of \$ 140,000 in profit. Therefore, by reducing home construction costs, you can make the business even more profitable.

Let’s recap. By simply applying the demand function, the marginal profit, and the maximum profit of the calculation, and nothing else, you were able to help your business increase its monthly profit from the ABC Home Community project by hundreds of thousands of dollars. Through a little negotiation with his construction suppliers and labor leaders, he was able to reduce his costs and through a simple readjustment of the cost equation (VS), you could quickly see that by cutting costs, you increased profits once 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.