Calculus has many real-world applications and applications in physics, computer science, economics, business, and medicine. I will briefly touch on some of these real estate applications and applications.
First, we start with the use of some examples of calculus, in particular, the development of real estate (i.e., new housing construction). Logically, a new home builder wants to make a profit after completing each home in a new home community. This builder will also have to maintain (hopefully) a positive cash flow during the construction of each house or at each stage of the development of the house. There are many factors that go into the calculation of profits. For example, we already know the profit formula: P = R - C , that is, the profit ( P ) is equal to income ( R ) minus the cost ( FROM ). Although this primary formula is very simple, there are many variables that can affect this formula. For example, at a cost ( FROM ), there are many different variable costs, such as the cost of building materials, labor costs, the cost of buying property before you buy, the cost of utilities and insurance premiums during the construction phase. These are several of the many costs associated with the above formula. Under the income ( R ), you can include variables such as the base price of the sale of the house, additional updates or additions to the house (security system, surround sound system, granite countertops, etc.). Simply incorporating all of these variables into itself can be a daunting task. However, it becomes even more difficult 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 of the areas where calculus comes into play.
Suppose, for example, last month we sold 50 houses with an average selling price of 500,000 dollars. Not taking into account other factors, our income ( R ) - the price (500,000 dollars) times x (50 houses sold), which is 25,000,000 US dollars. Suppose that the total construction cost of all 50 houses was $ 23,500,000; therefore profit ( P ) is 25 000 000 - 23 500 000 US dollars, which is 1 500 000 US dollars. Now that you know these numbers, your boss asked you to maximize your profits for the next month. How do you do it? What price can you set?
As a simple example, let us first calculate the marginal profit in terms of X building a house 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 determine that the demand for selling a home in this community
P = 1 000 000 US dollars - X / 10.
At $ 1,000,000, you know you will not sell any homes. Now the cost equation ( FROM ) is an
$ 300,000 + $ 18,000 X ($ 175,000 for fixed costs for materials and $ 10,000 for a house sold + $ 125,000 for fixed labor costs and $ 8,000 for a house).
From this you can calculate the marginal profit in terms of X (units sold), then use marginal profit to calculate the price we need to charge in order to maximize profit. Thus, the income
R = PV = (1,000,000 US dollars - X / 10) * ( X ) = 1,000,000 X - x ^ 2/10.
Therefore profit
P = R - C = (US $ 1,000,000 X - x ^ 2/10) - ($ 300,000 + $ 18,000) X ) = 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 marginal profit equal to zero and decide
982 000 - ( X / 5) = 0
X = 4910000.
We connect X back to the demand function and get the following:
P = $ 1,000,000 - (4,900,000) / 10 = $ 509,000.
Thus, the price we have to set in order to get the maximum profit for each house we sell must be $ 509,000. Next month, you will sell another 50 houses with a new pricing structure and make a profit of $ 450,000 compared with the previous month. Great job!
Now, next month, your boss will ask you, a community developer, to find a way to reduce the cost of housing. Before you know what the cost equation is ( FROM ) was:
$ 300,000 + $ 18,000 X ($ 175,000 for fixed costs for materials and $ 10,000 for a house sold + $ 125,000 for fixed labor costs and $ 8,000 for a house).
After insightful 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 labor costs to $ 110,000 and $ 7,000 per house. As a result, your cost equation ( FROM ) changed to
FROM = $ 260,000 + $ 16,000 X ,
Because of these changes, you will need to recalculate the basic profit.
P = R - C = (US $ 1,000,000 X - x ^ 2/10) - ($ 260,000 + $ 16,000) X ) = 984,000 X - ( x ^ 2/10) - $ 260,000.
From this we can calculate the new marginal profit, taking the derivative of the new profit, calculated
dP / dx = 984 000 - ( X / 5).
To calculate the maximum profit, we set marginal profit equal to zero and decide
984 000 - ( X / 5) = 0
X = 4920000.
We connect X back to the demand function and get the following:
P = 1 000 000 US dollars - (4920000) / 10 = 508 000 US dollars.
Thus, the price we have to set in order to get a new maximum profit for each house sold must be 508,000 US dollars. Now, although we are lowering the selling price from 509,000 dollars to 508,000 dollars, and we are still selling 50 units, as in the previous two months, our profit has still increased because we have reduced expenses by 140,000 US dollars. We can figure this out by calculating the difference between the first P = R - C and second P = R - C which contains the new cost equation.
first P = R - C = (US $ 1,000,000 X - x ^ 2/10) - ($ 300,000 + $ 18,000) X ) = 982,000 X - ( x ^ 2/10) - $ 300,000 = 48,799,750
second P = R - C = (US $ 1,000,000 X - x ^ 2/10) - ($ 260,000 + $ 16,000) X ) = 984,000 X - ( x ^ 2/10) - $ 260,000 = 48,939,750
Taking the second profit minus the first profit, you can see the difference (increase) in the amount of 140 000 dollars of profit. Thus, reducing the cost of housing construction, you can make the company even more profitable.
Let's go back. By simply applying the demand function, marginal profit and maximum profit from calculus, and nothing else, you couldn’t help your company increase its monthly income from the ABC Home Community project by hundreds of thousands of dollars. Thanks to small negotiations with your construction suppliers and labor managers, you were able to reduce your costs by simply adjusting the cost equation ( FROM ), you could quickly see that, by reducing costs, you increased your profits again, even after you set up your maximum profit by lowering your selling price by $ 1,000 per piece. This is an example of the miracle of calculus in relation to real problems.
