Problem 11 - Applied Calculus

Question

A company earns \(2 \%\) per month on its assets, paid continuously, and its expenses are paid out continuously at a rate of \(\$ 80,000\) per month. (a) Write a differential equation for the value, \(V\), of the company as a function of time, \(t\), in months. (b) What is the equilibrium solution for the differential equation? What is the significance of this value for the company? (c) Solve the differential equation found in part (a). (d) If the company has assets worth \(\$ 3\) million at time \(t=0\), what are its assets worth one year later?

Step-by-Step Solution

Answer

The assets are worth approximately \$3,213,400 after one year.

Step 1: Define the Problem

We need to set up a differential equation for the company's asset value \(V(t)\). The company earns at a rate of \(2\%\) per month, expressed in continuous compounding as \(0.02V\), and has continuous expenses of \(\$80,000\) per month.

Step 2: Write the Differential Equation

The change in the asset value \(V\) over time \(t\) can be represented by the differential equation \(\frac{dV}{dt} = 0.02V - 80000\). Here, \(0.02V\) represents the income, and the \(-80000\) represents the expenses.

Step 3: Find the Equilibrium Solution

The equilibrium solution occurs when \(\frac{dV}{dt} = 0\). Set \(0.02V - 80000 = 0\), solve for \(V\):\[ 0.02V = 80000 \]\[ V = \frac{80000}{0.02} = 4000000 \].The equilibrium solution is \(\$4,000,000\). This is the asset value where earnings and expenses are balanced, leading to no change in the asset value.

Step 4: Solve the Differential Equation

The equation \(\frac{dV}{dt} = 0.02V - 80000\) is linear and can be solved using an integrating factor \(\mu(t) = e^{0.02t}\). Multiply through by \(\mu(t)\):\[ e^{0.02t} \frac{dV}{dt} = 0.02 e^{0.02t} V - 80000 e^{0.02t} \].This gives:\[ \frac{d}{dt}(e^{0.02t} V) = 0.02 V e^{0.02t} \].Integrate both sides:\[ e^{0.02t} V = \int 80000 e^{0.02t} dt \].Solving, we find:\[ e^{0.02t} V = 4000000 e^{0.02t} + C \].Hence, \( V = 4000000 + Ce^{-0.02t} \).

Step 5: Determine the Constant Using Initial Condition

Use the initial condition \(V(0) = 3000000\):\[ 3000000 = 4000000 + C \right\]Solving for \(C\), we get \(C = -1000000\). Thus, the solution is \(V(t) = 4000000 - 1000000 e^{-0.02t}\).

Step 6: Calculate Value After One Year

Substitute \(t = 12\) months into \(V(t)\):\[ V(12) = 4000000 - 1000000 e^{-0.24} \].Calculating \(e^{-0.24} \approx 0.7866\), we get:\[ V(12) = 4000000 - 1000000 \times 0.7866 = 3213400 \].The asset value after one year is approximately \$3,213,400.

Key Concepts

Mathematical Modeling

Mathematical modeling is a powerful tool that helps us understand real-world phenomena using mathematical concepts. In this context, we use a differential equation to model the financial dynamics of a company. This equation reflects changes in asset value, taking into account continuous earnings and expenses.
To set up the model for this exercise, we identify key factors: the earnings rate, expenses, and asset value. Representing complex economic relationships as mathematical expressions allows for predictions and understanding of how these factors interact over time.
By creating this model using differential equations, we simulate the consistent flow of income and expenses, helping to predict the company’s financial trajectory under various scenarios. This is crucial for strategic planning and decision-making.

Continuous Compounding

In financial mathematics, continuous compounding refers to the process where interest is calculated and added to the principal continuously, as opposed to at fixed intervals. This allows the investment to grow exponentially over time.
For the company's assets, the 2% monthly earnings compounded continuously are described by the term \(0.02V\). This continuous growth is captured in the differential equation which models how earnings are incrementally added to the asset base every instant.
By using continuous compounding, we better capture real-world scenarios where changes happen consistently rather than at fixed periods. This method results in maximal interest accrual and more accurately reflects ongoing financial activities.

Equilibrium Solution

The equilibrium solution of a differential equation is the condition where the change in the variable of interest becomes zero. This means that at equilibrium, inflows equal outflows, resulting in a stable system.
For the given company, the equilibrium solution for the differential equation is when \frac{dV}{dt} = 0\. Solving \0.02V - 80000 = 0\, we find that \(V = 4000000\). This value represents the balance point where earnings precisely cover expenses.
Reaching an equilibrium is significant as it marks the point where the company’s assets neither grow nor shrink, indicating a financially sustainable state provided all other conditions remain constant.

Initial Value Problem

An initial value problem involves solving a differential equation with a given initial condition. This condition allows us to find a specific solution that matches the real-world scenario at a particular starting point.
In this case, the initial condition is that the company has \$3,000,000\ in assets at \(t=0\). This initial value helps determine the constant in the general solution of the differential equation. By substituting the initial value into the solution \(V(t) = 4000000 - 1000000 e^{-0.02t}\), we find specific behavior for \(V(t)\).
Solving the initial value problem allows for precise modeling of the company's asset value over time, enabling predictions and assessments of future financial status.