The population of a town increases at a rate proportional to the population at that time . If the population increases from 40 thousands to 60 thousands in 40 years, What will be the population in another 20 years ? = [ Given: ` sqrt(3/2) = 1.2247` ]

Answer

To solve the problem of population growth in a town, we will follow these steps:

### Step 1: Set up the differential equation
The population \( P(t) \) increases at a rate proportional to the population at that time. Therefore, we can express this relationship as:
\[ \frac{dP}{dt} = kP \]
where \( k \) is the proportionality constant.

### Step 2: Separate the variables and integrate
We can separate the variables and integrate both sides:
\[ \frac{dP}{P} = k \, dt \]
Integrating both sides gives:
\[ \ln P = kt + C \]
where \( C \) is the constant of integration.

### Step 3: Exponentiate to solve for \( P \)
Exponentiating both sides, we get:
\[ P = e^{kt + C} = e^C e^{kt} \]
Let \( P_0 = e^C \) (the initial population), so:
\[ P(t) = P_0 e^{kt} \]

### Step 4: Use the initial conditions to find \( k \)
We know that the population increases from 40,000 to 60,000 in 40 years. Thus, we can set up the equation:
\[ P(0) = 40000 \quad \text{and} \quad P(40) = 60000 \]
Using the equation:
\[ P(0) = P_0 e^{k \cdot 0} = P_0 = 40000 \]
Now substituting \( t = 40 \):
\[ 60000 = 40000 e^{40k} \]
Dividing both sides by 40000 gives:
\[ \frac{60000}{40000} = e^{40k} \implies 1.5 = e^{40k} \]
Taking the natural logarithm of both sides:
\[ \ln(1.5) = 40k \implies k = \frac{\ln(1.5)}{40} \]

### Step 5: Find the population after another 20 years (total 60 years)
Now we need to find \( P(60) \):
\[ P(60) = 40000 e^{60k} \]
Substituting \( k \):
\[ P(60) = 40000 e^{60 \cdot \frac{\ln(1.5)}{40}} = 40000 e^{\frac{3}{2} \ln(1.5)} = 40000 (1.5)^{\frac{3}{2}} \]
Calculating \( (1.5)^{\frac{3}{2}} \):
\[ (1.5)^{\frac{3}{2}} = \sqrt{(1.5)^3} = \sqrt{3.375} = \sqrt{\frac{27}{8}} = \frac{3\sqrt{3}}{4} \]
Using the approximation \( \sqrt{3/2} \approx 1.2247 \):
\[ (1.5)^{\frac{3}{2}} = (1.2247)^3 \approx 1.2247 \times 1.2247 \times 1.2247 \approx 1.836 \]
Thus:
\[ P(60) \approx 40000 \times 1.836 \approx 73440 \]

### Final Answer
The population of the town after another 20 years will be approximately **73,440**.

---