/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 An endowment is an investment ac... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 42

An endowment is an investment account in which the balance ideally remains constant and withdrawals are made on the interest earned by the account. Such an account may be modeled by the initial value problem \(B^{\prime}(t)=r B-m,\) for \(t \geq 0,\) with \(B(0)=B_{0} .\) The constant \(r>0\) reflects the annual interest rate, \(m>0\) is the annual rate of withdrawal, \(B_{0}\) is the initial balance in the account, and \(t\) is measured in years. a. Solve the initial value problem with \(r=0.05, m=\$ 1000 /\) year, and \(B_{0}=\$ 15,000 .\) Does the balance in the account increase or decrease? b. If \(r=0.05\) and \(B_{0}=\$ 50,000,\) what is the annual withdrawal rate \(m\) that ensures a constant balance in the account? What is the constant balance?

Short Answer

Expert verified
#Answer#: Part a: The solution to the initial value problem is \(B(t) = 20000 - 5000e^{0.05t}\). The balance in the account will decrease over time. Part b: The annual withdrawal rate m that ensures a constant balance is $2500 per year, and the constant balance will be $50,000.

Step by step solution

01

Part a: Solve the Initial Value Problem

To solve the given initial value problem, we can separate the variables and integrate. \(B^{\prime}(t)=0.05B(t)-1000\) for \(t \geq 0,\) and initial condition: \(B(0)=15000\). 鈧�. Separate the variables: \(\frac{dB}{dt} = 0.05B - 1000 \Rightarrow \frac{dB}{B-20000} = 0.05 dt\) 鈧�. Integrate both sides: \(\int \frac{1}{B-20000} dB = \int 0.05 dt\) \(\Rightarrow \ln|B-20000| = 0.05t + C_1\) 鈧�. Solve for B(t): \(B(t) = 20000 + Ce^{0.05t}\) 鈧�. Apply the initial condition \(B(0)=15000\): \(15000 = 20000 + Ce^{0(0.05)}\) \(C = -5000\) So, the solution to the initial value problem is \(B(t) = 20000 - 5000e^{0.05t}\).
02

Part a: Determine Increase or Decrease of Account Balance

To determine whether the balance in the account will increase or decrease, we can check the sign of \(B'(t)\) at \(t=0\): \(B'(t) = 0.05(20000 - 5000e^{0.05t}) - 1000\) \(B'(0) = 0.05(20000 - 5000) - 1000 = -500\) Since \(B'(0) < 0\), the balance in the account will decrease over time.
03

Part b: Annual Withdrawal Rate That Ensures Constant Balance

To maintain a constant balance, the rate of withdrawal must equal the amount of interest earned. Therefore, we have: \(r \times B_0 = m\) We're given r = 0.05 and \(B_0 = 50000\). Plugging in the values, we calculate the annual withdrawal rate m: \(m = 0.05 \times 50000 = 2500\) So, the annual withdrawal rate m is $2500 per year to maintain a constant balance in the account.
04

Part b: Calculate Constant Balance

Since we are maintaining a constant balance in the account, our constant balance is equal to the initial balance: Constant Balance = \(B_0 = 50000\) So, the constant balance in the account is $50,000.

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