The population of a city increases at a rate proportional to the population at that time. If the polultion of the city increase from 20 lakhs to 40 lakhs in 30 years, then after another 15 years, the polulation is

Options

A. `10sqrt2 lakhs`
B. `20sqrt2 lakhs`
C. `30sqrt2`
D. `40sqrqt2 lakhs`

Answer

To solve the problem, we will follow these steps:

### Step 1: Set up the differential equation
The problem states that the population \( P \) increases at a rate proportional to the population at that time. This can be expressed mathematically as:
\[ \frac{dP}{dt} = kP \]
where \( k \) is a constant of proportionality.

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

### Step 3: Solve for the constant of integration
To find \( C \), we need initial conditions. We know that at \( t = 0 \), \( P = 20 \) lakhs (20,00,000) and at \( t = 30 \), \( P = 40 \) lakhs (40,00,000).

Using the initial condition \( P(0) = 20 \):
\[ \ln(20) = k(0) + C \implies C = \ln(20) \]
So our equation becomes:
\[ \ln P = kt + \ln(20) \]

### Step 4: Use the second condition to find \( k \)
Now we use the second condition \( P(30) = 40 \):
\[ \ln(40) = k(30) + \ln(20) \]
Subtracting \( \ln(20) \) from both sides:
\[ \ln(40) - \ln(20) = 30k \]
Using the property of logarithms:
\[ \ln\left(\frac{40}{20}\right) = 30k \implies \ln(2) = 30k \]
Thus, we can solve for \( k \):
\[ k = \frac{\ln(2)}{30} \]

### Step 5: Find the population at \( t = 45 \)
Now we want to find the population at \( t = 45 \):
\[ \ln P = kt + \ln(20) \]
Substituting \( k \) and \( t = 45 \):
\[ \ln P = \frac{\ln(2)}{30}(45) + \ln(20) \]
Calculating \( \frac{45}{30} = 1.5 \):
\[ \ln P = 1.5 \ln(2) + \ln(20) \]
Using the property of logarithms:
\[ \ln P = \ln(20) + \ln(2^{1.5}) = \ln(20 \cdot 2^{1.5}) = \ln(20 \cdot \sqrt{8}) = \ln(20 \cdot 2\sqrt{2}) = \ln(40\sqrt{2}) \]

### Step 6: Exponentiate to find \( P \)
Exponentiating both sides gives:
\[ P = 40\sqrt{2} \]
Thus, the population after another 15 years (i.e., at \( t = 45 \) years) is \( 40\sqrt{2} \) lakhs.

### Final Answer
The population after another 15 years is \( 40\sqrt{2} \) lakhs.

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