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Chapter 1: Problem 11

How much can I borrow? The function in Example \(1.2\) can be rearranged to show the amount of money \(P=P(M, r, t)\), in dollars, that you can afford to borrow at a monthly interest rate of \(r\) (as a decimal) if you are able to make \(t\) monthly payments of \(M\) dollars: $$ P=M \times \frac{1}{r} \times\left(1-\frac{1}{(1+r)^{r}}\right) . $$ Suppose you can afford to pay \(\$ 350\) per month for 4 years. a. How much money can you afford to borrow for the purchase of a car if the prevailing monthly interest rate is \(0.75 \%\) ? (That is \(9 \%\) APR.) Express the answer in functional notation, and then calculate it. b. Suppose your car dealer can arrange a special monthly interest rate of \(0.25 \%\) (or \(3 \%\) APR). How much can you afford to borrow now? c. Even at \(3 \%\) APR you find yourself looking at a car you can't afford, and you consider extending the period during which you are willing to make payments to 5 years. How much can you afford to borrow under these conditions?

Short Answer

Expert verified
a) $14,204, b) $15,611, c) $18,991.

Step by step solution

01

Identify Variables and Constants for Part (a)

For part (a), we know that the monthly payment \(M = \$350\) and the payment period \(t\) is 4 years. Since there are 12 months per year, this means \(t = 4 \times 12 = 48\) months. The monthly interest rate \(r = 0.75\% = 0.0075\). We need to rearrange the given formula into functional notation and calculate \(P\).
02

Function Notation for Part (a)

Express the amount of money \(P\) you can afford to borrow in function notation based on the given formula. The function will be \(P(M, r, t) = M \times \frac{1}{r} \times \left(1 - \frac{1}{(1+r)^t}\right)\). Substitute the known values to calculate \(P\).
03

Calculate Amount for Part (a)

Substitute \(M = 350\), \(r = 0.0075\), and \(t = 48\) into the function:\[P(350, 0.0075, 48) = 350 \times \frac{1}{0.0075} \times \left(1 - \frac{1}{(1+0.0075)^{48}}\right)\]Calculate using a calculator:\[P \approx 350 \times 133.3333 \times (1 - 0.6954) = 350 \times 133.3333 \times 0.3046 \approx 14,204\].
04

Identify Variables and Constants for Part (b)

For part (b), the monthly interest rate \(r\) is reduced to \(0.25\% = 0.0025\), with \(M = 350\) and \(t = 48\) remaining the same. Use the same function \(P(M, r, t)\) and substitute the new interest rate to calculate \(P\).
05

Calculate Amount for Part (b)

Substitute \(M = 350\), \(r = 0.0025\), and \(t = 48\) into the function:\[P(350, 0.0025, 48) = 350 \times \frac{1}{0.0025} \times \left(1 - \frac{1}{(1+0.0025)^{48}}\right)\]Calculate using a calculator:\[P \approx 350 \times 400 \times (1 - 0.88849) = 350 \times 400 \times 0.11151 \approx 15,611\].
06

Identify Variables for Part (c)

For part (c), the monthly interest rate \(r = 0.0025\), the monthly payment \(M = 350\), and the payment period \(t\) is extended to 5 years, so \(t = 5 \times 12 = 60\). Use the same function to find \(P\).
07

Calculate Amount for Part (c)

Substitute \(M = 350\), \(r = 0.0025\), and \(t = 60\) into the function:\[P(350, 0.0025, 60) = 350 \times \frac{1}{0.0025} \times \left(1 - \frac{1}{(1+0.0025)^{60}}\right)\]Calculate using a calculator:\[P \approx 350 \times 400 \times (1 - 0.86435) = 350 \times 400 \times 0.13565 \approx 18,991\].

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Key Concepts

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

Loan Calculations
Loan calculations are essential for understanding how much money you can borrow based on your ability to make monthly payments. Let's say you want to buy a car and can pay \(350 monthly for it. To calculate how much you can borrow, we use a specific formula:
  • Amount you can borrow (\(P\))
  • Monthly payment (\(M\))
  • Monthly interest rate (\(r\))
  • Number of payments (\(t\))

The loan formula is:\[P(M, r, t) = M \times \frac{1}{r} \times \left(1 - \frac{1}{(1+r)^t}\right)\]This formula helps determine \(P\), or the total amount you can afford to borrow. In the problem example, your monthly payment is fixed at \)350, and the period of payment, \(t\), is derived by converting years into months. Adjust the formula inputs depending on the interest rate and duration your lender offers.
Interest Rates
Interest rates play a crucial role in determining how much you can borrow. When calculating loans, even a small change in the interest rate can have a significant effect. Let's explore how this works:
  • The monthly interest rate ($r$) is a decimal form of the annual percentage rate (APR) divided by 12. For example, a 9% APR means a 0.75% monthly rate.
  • At a higher interest rate, you can borrow less because more of your monthly payment goes towards paying interest.
  • When the interest rate is lower, such as 3% APR, represented as 0.25% monthly, you can borrow more because less of your payment goes toward interest.

In our example, you can borrow $14,204 at a 9% APR, but $15,611 at a reduced 3% APR, assuming all other loan terms remain the same.
Mathematical Modeling
Mathematical modeling involves creating equations to solve real-world problems like loan calculations. It involves using functions to predict outcomes based on different variables. Here's how it works with our loan example:
  • Identify variables: Monthly payment, interest rate, and payment period.
  • Use a function, like the one provided, to represent the relationship between these variables. This function helps us compute the result when variables change.
  • Test different scenarios: Changing either the interest rate or the number of payments affects how much you can borrow.

This approach allows you to anticipate how different loan terms change what you can afford. Expanding the payment period to 5 years increases the amount you can borrow - it's now $18,991 at a 3% APR. Mathematical modeling gives you the tools to make informed financial decisions.

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Most popular questions from this chapter

Research project: For this project you are to find and describe a function that is commonly used. Find a patient person whose job is interesting to you. Ask that person what types of calculations he or she makes. These calculations could range from how many bricks to order for building a wall to lifetime wages lost for a wrongful-injury settlement to how much insulin to inject. Be creative and persistentdon't settle for "I look it up in a table." Write a description, in words, of the function and how it is calculated. Then write a formula for the function, carefully identifying variables and units.

Reynolds number: The Reynolds number is very important in such fields as fluid flow and aerodynamics. In the case of a fluid flowing through a pipe, the Reynolds number \(R\) is given by $$ R=\frac{v d D}{\mu} $$ Here \(v\) is the velocity of the fluid in meters per second, \(d\) is the diameter of the pipe in meters, \(D\) is the density of the fluid in kilograms per cubic meter, and \(\mu\) is the viscosity of the fluid measured in newton- seconds per square meter. Generally, when the Reynolds number is above 2000 , the flow becomes turbulent, and rapid mixing occurs. \({ }^{10}\) When the Reynolds number is less than 2000 , the flow is streamline. Consider a fluid flowing through a pipe of diameter \(0.05\) meter at a velocity of \(0.2\) meter per second. a. If the fluid in the pipe is toluene, its viscosity is \(0.00059\) newton- seconds per square meter, and its density is 867 kilograms per cubic meter. Is the flow turbulent or streamline? b. If the toluene is replaced by glycerol, then the viscosity is \(1.49\) newton-seconds per square meter, and the density is \(1216.3\) kilograms per cubic meter. Is the glycerol flow turbulent or streamline?

Sales income: The following table shows the net monthly income \(N\) for a real estate agency as a function of the monthly real estate sales \(s\), both measured in dollars. $$ \begin{array}{|c|c|} \hline s=\text { Sales } & N=\text { Net income } \\ \hline 450,000 & 4000 \\ \hline 500,000 & 5500 \\ \hline 550,000 & 7000 \\ \hline 600,000 & 8500 \\ \hline \end{array} $$ a. Make a table showing, for each of the intervals in the table above, the average rate of change in \(N\). What pattern do you see? b. Use the average rate of change to estimate the net monthly income for monthly real estate sales of \(\$ 520,000\). In light of your answer to part a, how confident are you that your estimate is an accurate representation of the actual income? c. Would you expect \(N\) to have a limiting value? Be sure to explain your reasoning.

Equity in a home: When you purchase a home by securing a mortgage, the total paid toward the principal is your equity in the home. (Technically, the lending agency calculates your equity by subtracting the amount you still owe on your mortgage from the current value of your home, which may be higher or lower than your principal.) If your mortgage is for \(P\) dollars, and if the term of the mortgage is \(t\) months, then your equity \(E\), in dollars, after \(k\) monthly payments is given by $$ E=P \times \frac{(1+r)^{k}-1}{(1+r)^{t}-1} $$ Here \(r\) is the monthly interest rate as a decimal, with \(r=\mathrm{APR} / 12\).c. Find a formula that gives your equity after \(y\) years of payments. Suppose you have a home mortgage of \(\$ 400,000\) for 30 years at an APR of \(6 \%\). a. What is the monthly rate as a decimal? Round your answer to three decimal places. b. Express, using functional notation, your equity after 20 years of payments, and then calculate that value.

Head and aquifers: This is a continuation of Exercise 21. In underground water supplies such as aquifers, the water normally permeates some other medium such as sand or gravel. The head for such water is determined by first drilling a well down to the water source. When the well reaches the aquifer, pressure causes the water to rise in the well. The head is the height to which the water rises. In this setting, we get the pressure using Pressure \(=\) Density \(\times 9.8 \times\) Head \(.\) Here density is in kilograms per cubic meter, head is in meters, and pressure is in newtons per square meter. (One newton is about a quarter of a pound.) A sandy layer of soil has been contaminated with a dangerous fluid at a density of 1050 kilograms per cubic meter. Below the sand there is a rock layer that contains water at a density of 990 kilograms per cubic meter. This aquifer feeds a city water supply. Test wells show that the head in the sand is \(4.3\) meters, whereas the head in the rock is \(4.4\) meters. A liquid will flow from higher pressure to lower pressure. Is there a danger that the city water supply will be polluted by the material in the sand layer?
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