A cuboidal shaped godown with square base is to be constructed. Three times as much cost per square metre is incurred for constructing the roof as compared to the walls. Find the dimensions of the godown if it is to enclose a given volume and minimize the cost of constructing the roof and the walls.

Answer

To solve the problem of minimizing the cost of constructing a cuboidal godown with a square base while enclosing a given volume, we can follow these steps:

### Step 1: Define Variables
Let:
- \( L \) = length of one side of the square base
- \( H \) = height of the godown
- \( V \) = given volume of the godown

### Step 2: Volume Equation
The volume \( V \) of the cuboidal godown can be expressed as:
\[ V = L^2 \cdot H \]
From this, we can express \( H \) in terms of \( L \):
\[ H = \frac{V}{L^2} \]

### Step 3: Cost Function
Next, we need to calculate the cost of constructing the walls and the roof.

1. **Cost of the Roof**: The area of the roof is \( L^2 \) and the cost per square meter is \( 3k \) (where \( k \) is the cost per square meter for the walls). Therefore, the cost for the roof is:
\[ \text{Cost of Roof} = 3k \cdot L^2 \]

2. **Cost of the Walls**: The area of the walls is \( 4 \cdot (L \cdot H) = 4LH \) (since there are 4 walls). The cost for the walls is:
\[ \text{Cost of Walls} = k \cdot 4LH \]
Substituting \( H \) from the volume equation:
\[ \text{Cost of Walls} = k \cdot 4L \cdot \frac{V}{L^2} = \frac{4kV}{L} \]

3. **Total Cost**: The total cost \( C \) is the sum of the costs of the roof and the walls:
\[ C = 3kL^2 + \frac{4kV}{L} \]

### Step 4: Minimize the Cost Function
To minimize the cost, we need to differentiate \( C \) with respect to \( L \) and set the derivative to zero:
\[ \frac{dC}{dL} = 6kL - \frac{4kV}{L^2} \]
Setting the derivative equal to zero:
\[ 6kL - \frac{4kV}{L^2} = 0 \]
Rearranging gives:
\[ 6kL^3 = 4kV \]
Dividing by \( k \) (assuming \( k \neq 0 \)):
\[ 6L^3 = 4V \]
Thus,
\[ L^3 = \frac{2V}{3} \quad \Rightarrow \quad L = \left(\frac{2V}{3}\right)^{1/3} \]

### Step 5: Calculate Height \( H \)
Now substituting \( L \) back into the equation for \( H \):
\[ H = \frac{V}{L^2} = \frac{V}{\left(\frac{2V}{3}\right)^{2/3}} = \frac{V \cdot 3^{2/3}}{(2V)^{2/3}} = \frac{3^{2/3}}{2^{2/3}} \cdot V^{1/3} \]
Thus,
\[ H = \frac{3^{2/3}}{2^{2/3}} \cdot V^{1/3} \]

### Step 6: Final Dimensions
The dimensions of the godown that minimize the cost are:
- Length of the base \( L = \left(\frac{2V}{3}\right)^{1/3} \)
- Height \( H = \frac{3^{2/3}}{2^{2/3}} \cdot V^{1/3} \)