A square-based tank of capacity 250 cu m has to bedug out. The cost of land is Rs 50 per sq m. The cost of digging increases with the depth and for the whole tank the cost is Rs `400 xx (depth)^2`. Find the dimensions of the tank for the least total cost.

Answer

To find the dimensions of a square-based tank that minimizes the total cost of digging and land, we can follow these steps:

### Step 1: Define Variables
Let the side length of the square base be \( x \) meters and the height (depth) of the tank be \( h \) meters.

### Step 2: Volume Constraint
The volume of the tank is given as 250 cubic meters. The volume \( V \) of a square-based tank can be expressed as:
\[ V = x^2 \cdot h \]
Setting this equal to the given volume:
\[ x^2 \cdot h = 250 \quad \text{(1)} \]

### Step 3: Cost Function
The total cost \( C \) consists of two parts: the cost of the land and the cost of digging.

1. **Cost of Land**: The cost of land is Rs 50 per square meter. The area of the base is \( x^2 \), so the cost of land is:
\[ \text{Cost of Land} = 50 \cdot x^2 \]

2. **Cost of Digging**: The cost of digging is given as Rs \( 400h^2 \). Therefore, the total cost function can be expressed as:
\[ C = 50x^2 + 400h^2 \quad \text{(2)} \]

### Step 4: Substitute for \( x^2 \)
From equation (1), we can express \( x^2 \) in terms of \( h \):
\[ x^2 = \frac{250}{h} \]
Substituting this into the cost function (2):
\[ C = 50 \left(\frac{250}{h}\right) + 400h^2 \]
This simplifies to:
\[ C = \frac{12500}{h} + 400h^2 \quad \text{(3)} \]

### Step 5: Differentiate the Cost Function
To find the minimum cost, we need to differentiate \( C \) with respect to \( h \) and set the derivative equal to zero:
\[ \frac{dC}{dh} = -\frac{12500}{h^2} + 800h \]
Setting the derivative equal to zero:
\[ -\frac{12500}{h^2} + 800h = 0 \]
Rearranging gives:
\[ 800h^3 = 12500 \]
\[ h^3 = \frac{12500}{800} = 15.625 \]
Taking the cube root:
\[ h = \sqrt[3]{15.625} = 2.5 \text{ meters} \]

### Step 6: Find \( x \)
Now we can find \( x \) using equation (1):
\[ x^2 = \frac{250}{h} = \frac{250}{2.5} = 100 \]
Thus,
\[ x = \sqrt{100} = 10 \text{ meters} \]

### Final Dimensions
The dimensions of the tank for the least total cost are:
- Side length of the base \( x = 10 \) meters
- Height \( h = 2.5 \) meters

### Summary of Solution
- **Dimensions of the tank**: Length = Breadth = 10 meters, Height = 2.5 meters.