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Chapter 2: Problem 19

A savings and loan association estimates that the amount of money on deposit will be 1 million times the percentage rate of interest. For instance, a \(4 \%\) interest rate will generate $$\$ 4$$ million in deposits. If the savings and loan association can loan all the money it takes in at \(10 \%\) interest, what interest rate on deposits generates the greatest profit?

Short Answer

Expert verified
The interest rate on deposits that generates the greatest profit is 5%.

Step by step solution

01

Identify Variables

Let the interest rate on deposits be represented by \( r \) (expressed as a decimal).
02

Formulate the Deposit Function

The amount of money on deposit is given by \( 1,000,000 \times 100r = 1,000,000r \) dollars.
03

Determine the Total Revenue

The revenue from loans, which are distributed at a 10% interest rate, is \( 1,000,000r \times 0.10 = 100,000r \).
04

Calculate the Interest Paid on Deposits

The interest paid on deposits is \( 1,000,000r \times r = 1,000,000r^2 \).
05

Formulate the Profit Function

Profit, which is total revenue minus interest paid on deposits, can be written as: Profit = Total Revenue - Interest Paid on Deposits Profit = 100,000r - 1,000,000r^2.
06

Differentiate the Profit Function

To find the maximum profit, differentiate the profit function with respect to \( r \): \( P'(r) = 100,000 - 2,000,000r \).
07

Find the Critical Points

Set the derivative equal to zero to find critical points: \( 0 = 100,000 - 2,000,000r \) \( 2,000,000r = 100,000 \) \( r = \frac{100,000}{2,000,000} \) \( r = 0.05 \).
08

Check the Second Derivative

Differentiate the profit function's derivative to determine concavity: \( P''(r) = -2,000,000 \). Since \( P''(r) < 0 \), the function has a maximum at \( r = 0.05 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

derivatives
In calculus, a derivative represents how a function changes as its input changes. It is a key tool for understanding rates of change and is used extensively in optimization problems. To visualize, if you have a graph of a function, the derivative at any point is the slope of the tangent line to the curve at that point. In our problem, we used the derivative to determine how profit changes with respect to the interest rate on deposits. By differentiating the profit function, we derived the expression that helps us understand the rate at which the profit increases or decreases.
profit maximization
Profit maximization is a primary goal for many businesses. This concept refers to finding the point at which profit is at its greatest. Using calculus, specifically derivatives, we can determine the optimal conditions for maximum profit. In our exercise, we were tasked with finding the interest rate on deposits that maximizes profit. To do this, we set up a profit function: Profit = Total Revenue - Interest Paid on Deposits. We then used derivatives to understand how changing the interest rate would affect profits and found the rate that gives the highest possible profit.
critical points
Critical points are where the derivative of a function is zero or undefined. They are potential points for local maxima, minima, or saddle points. In optimization problems, finding these points helps us identify where the function changes direction, potentially highlighting maximum or minimum values. In our problem, once we differentiated the profit function, we set the derivative equal to zero to find the interest rate that might give us the highest profit. The calculation led us to the critical point at which the interest rate is 0.05.
second derivative test
The second derivative test helps confirm whether a critical point is a local maximum, minimum, or a saddle point. It involves taking the second derivative of the function and evaluating it at the critical points. If the second derivative is positive at a critical point, the function has a local minimum there. If it is negative, there is a local maximum. In our exercise, we calculated the second derivative of the profit function and found it to be negative. Therefore, the critical point at an interest rate of 0.05 is indeed a maximum, confirming that setting this interest rate maximizes the profit.

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