Problem 34 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Bernardo is a real estate developer. He estimates that if 60 luxury houses are built in a certain area, the average profit will be \(\$ 47,500\) per house. The average profit will decrease by \(\$ 500\) per house for each additional house built in the area. How many houses should Bernardo build to maximize the total profit? (Remember, the answer must be an integer.)

Step-by-Step Solution

Answer

Bernardo should build 77 or 78 houses to maximize profit.

Step 1: Define the Variables

Let the number of houses be represented by the variable \(x\).

Step 2: Express Profit per House

The profit per house decreases by \(\$ 500\) for each additional house. Thus, the profit per house can be expressed as: \[ P = 47500 - 500(x - 60) \]

Step 3: Express Total Profit

The total profit is the product of the number of houses and the profit per house. Therefore, the total profit function \(T(x)\) is: \[ T(x) = x \cdot (47500 - 500(x - 60)) \]

Step 4: Simplify the Total Profit Function

Simplify the total profit function \(T(x)\): \[ T(x) = x \cdot (47500 - 500x + 30000) \]\[ T(x) = 77500x - 500x^2 \]

Step 5: Find the Maximum of the Quadratic Function

The total profit function is a quadratic equation in the form \( ax^2 + bx + c \) with \(a = -500\) and \(b = 77500\). The maximum value of this quadratic function occurs at \( x = -\frac{b}{2a} \). Calculate this value:\[ x = -\frac{77500}{2(-500)} \]\[ x = 77.5 \]

Step 6: Adjust to Integer

Since the number of houses must be an integer, consider the two closest integers to 77.5: 77 and 78. Calculate the total profit for both:

Step 7: Calculate Total Profit for 77 Houses

For 77 houses: \[ T(77) = 77500(77) - 500(77)^2 \]\[ T(77) = 5967500 - 500(5929) \]\[ T(77) = 5967500 - 2964500 \]\[ T(77) = 3003000 \]

Step 8: Calculate Total Profit for 78 Houses

For 78 houses: \[ T(78) = 77500(78) - 500(78)^2 \]\[ T(78) = 6045000 - 500(6084) \]\[ T(78) = 6045000 - 3042000 \]\[ T(78) = 3003000 \]

Step 9: Conclusion

Since the total profit is the same for both 77 and 78 houses, Bernardo can build either 77 or 78 houses to maximize his total profit.

Key Concepts

Quadratic Functions

Quadratic functions are a type of polynomial function where the highest degree of the variable is 2. They are generally given in the form:

• \(ax^2 + bx + c\)

Here, ‘a’ is the coefficient of the quadratic term, ‘b’ is the coefficient of the linear term, and ‘c’ is a constant term. Quadratic functions create a parabolic curve when graphed, which can either open upwards or downwards.Understanding the shape of the parabola is crucial because it helps us to determine whether the quadratic function has a maximum or minimum value.
For example, in the problem of maximizing profit, the total profit function was simplified to:

• \(T(x) = -500x^2 + 77500x\)

The negative coefficient of the \(x^2\) term (-500) tells us that the parabola opens downwards, indicating that the quadratic function has a maximum value.This means that the maximum profit Bernardo can gain will be at the vertex of the parabola. The vertex can be found using the formula:

• \( x = -\frac{b}{2a}\)

which gives us the value of 'x' that maximizes the total profit.

Profit Maximization

Profit maximization is a crucial concept in business and economics. It involves determining the optimal level of output or input where profits are highest. In the context of quadratic functions, profit maximization often entails using calculus techniques to find the maximum point of a parabola.
For Bernardo's real estate development, the profit per house is \(47500\), but it decreases by \(500\) for each additional house built. The total profit is found by multiplying the number of houses (\( x \)) by the profit per house (\( P \)).

• You end up with the total profit function \( T(x) = x \times (47500 - 500(x - 60)) \).
• This simplifies to \( T(x) = 77500x - 500x^2 \), a quadratic function.

To find the number of houses that maximize the profit, you use the vertex formula for quadratics. Calculating the vertex gives you \( x = 77.5 \).
Since the number of houses must be an integer, you check both 77 and 78 houses. The total profit for both is the same, meaning Bernardo can build either 77 or 78 houses to maximize his profit.

Derivatives

Derivatives are a fundamental tool in calculus that measure how a function changes as its input changes. They are particularly useful for finding maximum and minimum values of functions, which is essential for solving optimization problems.
In the case of maximizing profit with quadratic functions, the derivative helps find the critical points where the function could have a maximum or minimum.Let's break it down:

• Start by taking the derivative of the total profit function \(T(x) = 77500x - 500x^2\).
• The derivative, denoted \(T'(x)\), will tell you how the total profit changes as the number of houses changes.
• The derivative is \(T'(x) = 77500 - 1000x \).

To find the critical points, set the derivative to zero:

• \(77500 - 1000x = 0\)
• Solve for \(x\): \(x = \frac{77500}{1000} = 77.5\)

This shows that the maximum profit occurs when around 77.5 houses are built. Since we can't build a fraction of a house, we check both 77 and 78 to confirm the best integer value.
By using derivatives, you can systematically analyze how changes in variables affect the overall outcome, which is invaluable for optimization.