The volume of a closed square based rectangular box is 1000 cubic metre. The cost of constructing the base is 15 paise per square metre and the cost of constructing the top is 25 paise per square metre. The cost of constructing its sides is 20 paise per square metre and the cost of constructing the box is Rs. 3.Find the dimensions of box for minimum cost of construction.

Answer

To find the dimensions of a closed square-based rectangular box that minimizes the cost of construction while maintaining a volume of 1000 cubic meters, we can follow these steps:

### Step 1: Define Variables
Let:
- \( l \) = length of the base (in meters)
- \( h \) = height of the box (in meters)

Since the box has a square base, the breadth \( b \) is equal to the length \( l \).

### Step 2: Volume Constraint
The volume \( V \) of the box is given by:
\[ V = l^2 \cdot h = 1000 \text{ cubic meters} \]
From this, we can express \( h \) in terms of \( l \):
\[ h = \frac{1000}{l^2} \]

### Step 3: Cost Function
The cost of constructing the box includes the cost of the base, top, and sides. The costs are as follows:
- Cost of the base (15 paise/m²):
\[ C_{\text{base}} = 15 \cdot l^2 \]
- Cost of the top (25 paise/m²):
\[ C_{\text{top}} = 25 \cdot l^2 \]
- Cost of the sides (20 paise/m²): The box has 4 sides, each with area \( l \cdot h \):
\[ C_{\text{sides}} = 4 \cdot (20 \cdot l \cdot h) = 80 \cdot l \cdot h \]

Combining these, the total cost \( C \) (in paise) is:
\[ C = C_{\text{base}} + C_{\text{top}} + C_{\text{sides}} = 15l^2 + 25l^2 + 80lh \]
\[ C = 40l^2 + 80l \cdot \frac{1000}{l^2} \]
\[ C = 40l^2 + \frac{80000}{l} \]

### Step 4: Minimize the Cost Function
To find the minimum cost, we differentiate \( C \) with respect to \( l \) and set the derivative equal to zero:
\[ \frac{dC}{dl} = 80l - \frac{80000}{l^2} \]
Setting the derivative equal to zero:
\[ 80l - \frac{80000}{l^2} = 0 \]
\[ 80l^3 = 80000 \]
\[ l^3 = 1000 \]
\[ l = 10 \text{ meters} \]

### Step 5: Calculate Height
Now, substituting \( l = 10 \) back into the equation for \( h \):
\[ h = \frac{1000}{l^2} = \frac{1000}{10^2} = \frac{1000}{100} = 10 \text{ meters} \]

### Step 6: Conclusion
The dimensions of the box for minimum cost are:
- Length \( l = 10 \) meters
- Height \( h = 10 \) meters
- Breadth \( b = 10 \) meters