Problem 19 - Precalculus: Mathematics for Calculus

Question

Mortgage Dr. Gupta is considering a 30-year mortgage at \(6 \%\) interest. She can make payments of \(\$ 3500\) a month. What size loan can she afford?

Step-by-Step Solution

Answer

Dr. Gupta can afford a loan of approximately $590,755.

Step 1: Identify the Monthly Interest Rate

We are given an annual interest rate of \(6\%\). To find the monthly interest rate, we need to divide the annual rate by 12 months. Therefore, the monthly interest rate is \(\frac{6\%}{12} = 0.5\%\). In decimal form, this is \(0.005\).

Step 2: Determine the Number of Payments

Dr. Gupta is considering a 30-year mortgage. Since payments are made monthly, the number of payments \(n\) is calculated by multiplying the number of years by 12 months per year: 30 years \(\times\) 12 months/year \(= 360\) payments.

Step 3: Use the Monthly Payment Formula

The monthly mortgage payment formula is given by: \[ M = \frac{P \times r \times (1 + r)^n}{(1 + r)^n - 1} \] where \(M\) is the monthly payment (\$3500), \(P\) is the principal (the loan amount, which we want to find), \(r\) is the monthly interest rate (0.005), and \(n\) is the number of payments (360).

Step 4: Rearrange Formula to Solve for Loan Amount

Rearrange the formula to solve for \(P\):\[ P = \frac{M \times ((1 + r)^n - 1)}{r \times (1 + r)^n} \]Substitute the known values into the formula: \[ P = \frac{3500 \times ((1 + 0.005)^{360} - 1)}{0.005 \times (1 + 0.005)^{360}} \]

Step 5: Calculate the Loan Amount

Calculate \((1 + 0.005)^{360}\) using a calculator:\((1.005)^{360} \approx 6.0225755707\).Substitute this back into the formula:\[ P = \frac{3500 \times (6.0225755707 - 1)}{0.005 \times 6.0225755707} \]\[ P \approx \frac{3500 \times 5.0225755707}{0.0301128} \approx 590,755.122 \]Therefore, the loan amount Dr. Gupta can afford is approximately \$590,755.

Key Concepts

Monthly Interest Rate

When you're considering a mortgage or any loan, it's important to understand what a monthly interest rate is and how to calculate it. The interest rate is typically given as an annual percentage rate (APR), but since payments are typically made monthly, you'll need a monthly rate for calculations. To convert the annual rate to a monthly rate, you simply divide the annual interest rate by 12, because there are 12 months in a year. For example, if you have an annual rate of 6%, like Dr. Gupta from our exercise, you'd calculate the monthly rate as follows:

• Annual rate: 6%
• Monthly rate: \( \frac{6\%}{12} = 0.5\% \)

Expressing this in decimal form makes it useful for plugging into financial formulas. So, 0.5% is represented as 0.005 in decimal form.

Loan Amount

The loan amount, sometimes referred to as the principal, is essentially the size of the loan you are seeking or able to obtain. Knowing what loan amount you can afford is crucial when considering a mortgage. The loan amount will affect your monthly payments and the total amount of interest you pay over time. In our scenario, Dr. Gupta needs to determine what size loan she can afford if she can pay $3,500 a month. The loan amount depends on several factors including:

• The monthly payment you can afford (here, $3,500)
• The monthly interest rate (0.005 in decimal form)
• The total number of payments (360 for a 30-year mortgage)

Using the monthly payment formula and solving for the principal gives her the maximum loan amount she can afford.

Monthly Payment Formula

Understanding the monthly payment formula is key to determining the loan amount for a mortgage. The formula is structured to calculate the amount you need to pay each month given a loan amount, interest rate, and number of payments. It’s represented as:\[M = \frac{P \times r \times (1 + r)^n}{(1 + r)^n - 1}\]Here, \(M\) stands for monthly payment, \(P\) is the principal (the loan amount), \(r\) is the monthly interest rate, and \(n\) is the number of monthly payments. For Dr. Gupta:

• \(M = 3500\)
• \(r = 0.005\)
• \(n = 360\)

To find the loan amount \(P\), you rearrange the formula. Solving for \(P\) involves:\[P = \frac{M \times ((1 + r)^n - 1)}{r \times (1 + r)^n}\]By substituting Dr. Gupta's details into the formula, she finds out that she can afford a loan of approximately $590,755.