A city building code requires that the area of the windows must be at least `(1)/(8)` of the area of the walls and roots of all new building. The daily heating cost of a new building is ₹ 3 per square meter of window area and ₹ 1 per square metre of wall and roof area. To the nearest square metre of wall and roof area. To the nearest square metre, what is the largest surface area a new building can have if its daily heating cost cannot exceed ₹ 1,000 ?

Answer

To solve the problem step by step, we will define the variables, set up the equations based on the given conditions, and then solve for the maximum surface area of the building.

### Step 1: Define the Variables
Let:
- \( A \) = Area of the walls and roof (in square meters)
- \( W \) = Area of the windows (in square meters)

According to the problem, the area of the windows must be at least \( \frac{1}{8} \) of the area of the walls and roof. Therefore, we can express this relationship as:
\[ W = \frac{A}{8} \]

### Step 2: Set Up the Cost Equation
The daily heating cost is given as:
- ₹ 3 per square meter for the window area
- ₹ 1 per square meter for the wall and roof area

The total heating cost \( C \) can be expressed as:
\[ C = 1 \cdot A + 3 \cdot W \]
Substituting \( W \) from the previous step:
\[ C = A + 3 \cdot \frac{A}{8} \]
This simplifies to:
\[ C = A + \frac{3A}{8} = \frac{8A}{8} + \frac{3A}{8} = \frac{11A}{8} \]

### Step 3: Set Up the Inequality
We know that the daily heating cost cannot exceed ₹ 1,000. Therefore, we set up the inequality:
\[ \frac{11A}{8} \leq 1000 \]

### Step 4: Solve for \( A \)
To isolate \( A \), we multiply both sides of the inequality by 8:
\[ 11A \leq 8000 \]
Now, divide both sides by 11:
\[ A \leq \frac{8000}{11} \approx 727.27 \]

### Step 5: Calculate the Maximum Window Area
Now we can calculate the maximum window area \( W \):
\[ W = \frac{A}{8} \leq \frac{727.27}{8} \approx 90.91 \]

### Step 6: Calculate the Total Surface Area
The total surface area \( S \) of the building is the sum of the wall and roof area and the window area:
\[ S = A + W = A + \frac{A}{8} = \frac{8A}{8} + \frac{A}{8} = \frac{9A}{8} \]
Substituting the maximum value of \( A \):
\[ S \leq \frac{9 \cdot 727.27}{8} \approx \frac{6545.43}{8} \approx 818.18 \]

### Conclusion
To the nearest square meter, the largest surface area \( S \) a new building can have is:
\[ \boxed{818} \text{ square meters} \]