Olivia is looking to take out a 30-year mortgage from a bank offering an annual interest rate of $3 \%$, compounded monthly Using the formula below, determine the maximum amount Olivia can borrow, to the nearest dollar, if the highest monthly payment she can afford is $\$ 1,975$. \[ M=\frac{\operatorname{Pr}(1+r)^{n}}{(1+r)^{n}-1} \] $M=$ the monthly payment $P=$ the amount borrowed $r=$ the interest rate per month $n=$ the number of payments

Step 1 ：Given that Olivia is looking to take out a 30-year mortgage from a bank offering an annual interest rate of 3%, compounded monthly. The highest monthly payment she can afford is $1,975. We are asked to determine the maximum amount Olivia can borrow.

Step 2 ：We use the formula for the monthly payment of a loan: \[M=\frac{\operatorname{Pr}(1+r)^{n}}{(1+r)^{n}-1}\] where \(M\) is the monthly payment, \(P\) is the amount borrowed, \(r\) is the interest rate per month, and \(n\) is the number of payments.

Step 3 ：We know the values for \(M\), \(r\), and \(n\). \(M\) is the monthly payment Olivia can afford, which is $1,975. \(r\) is the monthly interest rate, which is the annual interest rate divided by 12. The annual interest rate is 3%, so \(r\) is \(\frac{3}{100}/12 = 0.0025\). \(n\) is the number of payments, which is the number of months in 30 years, or \(30*12 = 360\).

Step 4 ：We substitute these values into the formula and solve for \(P\): \[1975 = \frac{P*0.0025*(1+0.0025)^{360}}{(1+0.0025)^{360}-1}\]

Step 5 ：Solving the equation, we find that the maximum amount Olivia can borrow, to the nearest dollar, is \(\boxed{468,449}\).