Problem 33 - A Graphical Approach to Precalculus with Limits

Question

Solve each problem. The manager of an 80-unit apartment complex knows from experience that at a rent of \(\$ 400\) per month, all units will be rented. However, for each increase of \(\$ 20\) in rent, he can expect one unit to be vacated. Let \(x\) represent the number of \(\$ 20\) increases over \(\$ 400\). (a) Express, in terms of \(x\), the number of apartments that will be rented if \(x\) increases of \(\$ 20\) are made. (For example, with three such increases, the number of apartments rented will be \(80-3=77\).) (b) Express the rent per apartment if \(x\) increases of \(\$ 20\) are made. (For example, if he increases rent by \(\mathrm{S} 60=3 \times \$ 20,\) the rent per apartment is given by \(400+3(20)=\$ 460 .)\) (c) Determine a revenue function \(R\) in terms of \(x\) that will give the revenue generated as a function of the number of \(\$ 20\) increases. (d) For what number of increases will the revenue be \(\$ 37,500 ?\) (e) What rent should he charge in order to achieve the maximum revenue?

Step-by-Step Solution

Answer

(a) \( 80-x \). (b) \( 400+20x \). (c) \( R(x) = 32000+1600x-20x^2 \). (d) Solve \( -20x^2+1600x+32000=37500 \). (e) Charge \( \$1200 \).

Step 1: Define the number of rented apartments

The problem states that for each \(20 increase, one unit will be vacated. Thus, if there are \( x \) increases of \)20, the number of rented apartments can be expressed as \( 80 - x \).

Step 2: Define the rent per apartment

The initial rent is \(400. For each \( x \) increases of \)20, the rent per apartment will be \( 400 + 20x \).

Step 3: Develop the revenue function

Revenue is calculated by multiplying the number of rented apartments by the rent per apartment. So, the revenue function \( R(x) \) is given by \( R(x) = (80 - x)(400 + 20x) \).

Step 4: Expand the revenue function

To simplify \( R(x) \), distribute and combine like terms:\[ R(x) = (80 - x)(400 + 20x) = 32000 + 1600x - 20x^2 \].

Step 5: Find the maximum revenue

The revenue function is a quadratic equation in the form \( -20x^2 + 1600x + 32000 \). The maximum value of a quadratic function \( ax^2 + bx + c \) with negative \( a \) occurs at \( x = -\frac{b}{2a} \). Calculate:\[ x = -\frac{1600}{2(-20)} = 40 \].

Step 6: Calculate the rent for maximum revenue

Substitute \( x = 40 \) back into the rent equation from Step 2:\[ 400 + 20 \times 40 = 1200 \].

Step 7: Solve for specific revenue

Set the revenue function equal to \(37,500 and solve for \( x \):\[ -20x^2 + 1600x + 32000 = 37500 \].Simplify and solve the quadratic:\[ -20x^2 + 1600x - 5500 = 0 \].Using the quadratic formula yields solutions for \( x \) which approximately may indicate specific number of increases leading to \)37,500 revenue.

Key Concepts

Revenue Optimization

Revenue optimization is about finding the best price strategy to maximize income. In our example, the apartment complex manager can adjust rent by $20 increments. For each $20 increase, one tenant will leave, which changes the number of occupied units. We explore how varying prices and units affect total revenue. Revenue is defined as the total money earned, which is the rent per apartment multiplied by the number of occupied apartments. The central idea behind revenue optimization here is to find the rent that generates the highest total revenue. This involves using algebra to create and solve a revenue function, a mathematical expression showing how revenue depends on rent changes. In the scenario, the revenue function is quadratic due to its maximum point, indicating rent's effect over time isn't linear. Calculating the vertex of the quadratic function helps determine the peak revenue point, which tells the manager just how much to charge for maximum profit.

Algebraic Expressions

Algebraic expressions are combinations of numbers and letters forming a mathematical relationship. They help in transforming verbal descriptions into mathematical formulas. In the context of the given problem, algebraic expressions describe how changes in rent impact occupancy and revenue.For example, to express the number of rented apartments when the manager increases rent by \(20 increments, we use the expression:

• Number of rented apartments: \(80 - x\), where \(x\) represents \)20 increases.

Similarly, the rent for each apartment after increments is represented by:

• Rent per apartment: \(400 + 20x\), reflecting how rent rises with every increment.

These expressions are then used to formulate the revenue function, which is a cumulative representation of potential income based on varying rent levels.

Revenue Function

The revenue function is a key tool to determine total income based on price changes. In this case, it's represented as a quadratic function:\[ R(x) = (80 - x)(400 + 20x) \]This equation derives from multiplying the number of units rented by the rent price, showing the combined effects of both variables on revenue.Let's expand this equation:\[ R(x) = 32000 + 1600x - 20x^2 \]Here, the term \(-20x^2\) indicates the revenue will eventually decrease with too many rent hikes, as it affects occupancy.The highest point, or maximum revenue, is found using the formula for the vertex of a parabola, specifically for the form \( ax^2 + bx + c \). This helps calculate the exact rent price for maximum profit and shows the importance of setting a strategic rent increase level to optimize revenue. The conclusion brings together how algebra and functions work together to drive financial decisions.