91Ó°ÊÓ

In finance, the interest rate is the percentage charged on the total amount of money deposited or borrowed. It represents the cost of borrowing money or the gain from lending it. Banks and savings associations often set different interest rates for deposits and loans.

For example, an interest rate of 5% means that for every \(100 deposited, the depositor earns \)5 annually. In our optimization problem, we denote the interest rate on deposits by the variable, \( r \), where \( r \) is expressed as a percentage. If \( r \) is 4%, then the deposits amount to \( 1,000,000 \times 0.04 = 4,000,000 \) dollars.

Understanding how interest rate impacts deposits and loans is crucial in solving optimization problems like this. The relationship between interest rate and generated deposits was the basis for setting up our problem.

A profit function analyzes the earnings of a business after accounting for all expenses. In our case, the profit function calculates the savings and loan association's profits based on interest rates.

We first determine the revenue, which comes from loaning out the deposits at 10% interest. Then, we subtract the expenses, namely the interest paid on the original deposits. As we defined our variables:

• Total deposits = \( 1,000,000r \)
• Interest earned from loans = \( 0.10 \times 1,000,000r \)
• Interest paid on deposits = \( 10,000r^2 \)

The profit function \( P(r) \) is therefore: \( P(r) = 100,000r - 10,000r^2 \).

This function helps us understand how different interest rates affect profitability. Higher interest rates draw more deposits, increasing loan profits but also boosting payout costs on these deposits. Balancing these factors is necessary to pinpoint the rate that maximizes profit.

Derivatives in calculus measure how a function changes as its input changes. They are crucial in optimization problems because they help identify maxima and minima.

For our profit function \( P(r) = 100,000r - 10,000r^2 \), we take the derivative, \( P'(r) \), to find the rate of change of profit regarding the interest rate. The first derivative tells us where the profit function's slope is zero, a point that could be a maximum or minimum.

We compute it as:
\( P'(r) = 100,000 - 20,000r \).

Setting \( P'(r) = 0 \) allows us to find critical points. Solving \( 100,000 - 20,000r = 0 \) gives us \( r = 5 \). This rate needs further validation as a maximum, which we achieve by checking the second derivative.

The second derivative \( P''(r) = -20,000 \) being negative confirms the critical point is a maximum, as the function opens downward. Thus, finding and interpreting derivatives is key to solving optimization problems effectively.

Maximizing profit involves finding the best conditions under which a business earns the highest returns. In this savings and loan association scenario, the goal was to find the interest rate that results in the highest profit.

The steps started with defining our profit function, \( P(r) = 100,000r - 10,000r^2 \), and then finding its derivative to locate the critical point. Solving \(P'(r) = 0\) for \( r \) led us to an interest rate of 5%.

Next, we confirmed this was a maximum by ensuring the second derivative was negative. The mathematical analysis showed:

• First derivative: \( P'(r) = 100,000 - 20,000r \)
• Critical point at \( r = 5 \)
• Second derivative: \( P''(r) = -20,000 \)
• Negative second derivative means a maximum at \( r = 5 \)

This optimization demonstrates how calculus principles help businesses make informed decisions. By using derivatives, we determined that a 5% interest rate on deposits yields the maximum profit for the association.