Problem 42 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Lynbrook West, an apartment complex, has 100 two-bedroom units. The monthly profit (in dollars) realized from renting out \(x\) apartments is given by $$ P(x)=-10 x^{2}+1760 x-50,000 $$ To maximize the monthly rental profit, how many units should be rented out? What is the maximum monthly profit realizable?

Step-by-Step Solution

Answer

To maximize the monthly rental profit, 88 units should be rented out, and the maximum monthly profit realizable is $77,440.

Step 1: Identify the quadratic function

In this case, the quadratic function is given by: \[ P(x) = -10x^2 + 1760x - 50000 \]

Step 2: Find the x-coordinate of the vertex

The x-coordinate of the vertex of a quadratic function in the form \( ax^2 + bx + c \) is given by: \[ x_{vertex} = \frac{-b}{2a} \] where a = -10 and b = 1760. Plugging these values into the formula, we get: \[ x_{vertex} = \frac{-1760}{2(-10)} \]

Step 3: Calculate the x-coordinate

By solving the expression, we find the x-coordinate of the vertex: \[ x_{vertex} = \frac{-1760}{-20} = 88 \] So to maximize the monthly rental profit, 88 units should be rented out.

Step 4: Find the y-coordinate of the vertex (maximum profit)

Now, we will plug in the x-coordinate of the vertex into the original profit function to find the corresponding maximum profit: \[ P(88) = -10(88)^2 + 1760(88) - 50000 \]

Step 5: Calculate the maximum profit

By solving this expression, we find the maximum monthly profit realizable: \[ P(88) = -10(7744) + 154880 - 50000 \] \[ P(88) = -77440 + 154880 - 50000 \] \[ P(88) = 77,440 \] The maximum monthly profit realizable is $77,440. In conclusion, to maximize the monthly rental profit, 88 units should be rented out, and the maximum monthly profit realizable is $77,440.

Key Concepts

Quadratic Functions

Quadratic functions play a crucial role in mathematics and real-world applications. They are typically represented by the equation \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. Quadratic functions yield a parabolic graph. The highest or lowest point of this parabola, depending on the direction, is called the vertex.

• If \( a \) is positive, the parabola opens upwards, and the vertex represents the minimum point.
• If \( a \) is negative, the parabola opens downwards, indicating that the vertex is the maximum point.

Understanding quadratic functions helps in various contexts, such as physics, engineering, and economics. Applied correctly, they can help maximize profits, optimize designs, or predict projectile paths. Quadratic equations' distinct advantage is their ability to model a multitude of natural phenomena and financial models.

Vertex of a Parabola

The vertex of a parabola is key to understanding quadratic functions. It represents the point where the parabola reaches its minimum or maximum value, according to its direction.
The vertex provides important information:

• The \( x \)-coordinate of the vertex can be found using \( \frac{-b}{2a} \) in any quadratic equation \( ax^2 + bx + c \).
• The \( y \)-coordinate of the vertex shows the value of the quadratic function at the vertex.
• In business or physics, the vertex can indicate optimal values, like maximum profit or trajectory height.

In the given problem, the vertex is at \( x = 88 \), meaning 88 units should be rented to maximize profit. This concept is crucial for solving quadratic optimization problems in various disciplines.

Profit Maximization

Profit maximization is a common goal in economics and business operations. It involves determining the level of output that generates the highest possible profit. The exercise uses a quadratic function to model profit, a strategy frequently employed in economics due to its predictive accuracy.
In the problem:

• The profit function \( P(x) = -10x^2 + 1760x - 50000 \) dictates how profit changes with rented units.
• Finding the vertex gives the profit-maximizing quantity, \( x = 88 \), hence rent out 88 apartments for optimal profit.
• Calculating \( P(88) \) reveals the maximum monthly profit achievable: $77,440.

Utilizing quadratic optimization helps businesses make data-driven decisions, balancing resources effectively to achieve financial goals. Thus, in applied mathematics, such models guide strategic decision-making processes.

Applied Mathematics

Applied mathematics involves mathematical methods used in practical fields like engineering, finance, and economics. It connects theoretical knowledge with real-world issues, providing solutions to optimize, predict, or control processes. By using quadratic functions, applied mathematics enables efficient problem-solving.
Real-world applications include:

• Predicting profits or costs in economics.
• Optimizing designs in engineering, such as structural forms or materials planning.
• Modeling natural events in physics, like projectile motion or population growth.

In the exercise, applied mathematics helps solve a profit-maximization problem, showcasing how complex concepts like quadratic optimization can address everyday business challenges. The blend of mathematics with practical applications allows for informed decision-making and strategic planning across various fields.