It is required to construct a big rectangular hall that can accommodate 400 people with ` 25 m^(3)` space for each person . The height of the wall has been fixed at 10 m and the total inner surface area of the walls must be `1300 m^(2)` . What is the length and breadth of the hall ( in metres ) ?

Options

A. `30,20`
B. `45,20`
C. `40,25`
D. `35,30`

Answer

To solve the problem of constructing a rectangular hall that can accommodate 400 people with specific volume and surface area requirements, we can follow these steps:

### Step 1: Calculate the Total Volume Required
Each person requires 25 m³ of space, and there are 400 people. Thus, the total volume \( V \) required is:
\[ V = \text{Number of people} \times \text{Volume per person} = 400 \times 25 = 10000 \, m^3 \]

### Step 2: Relate Volume to Dimensions
The volume of a rectangular hall can be expressed as:
\[ V = L \times B \times H \]
where \( L \) is the length, \( B \) is the breadth, and \( H \) is the height. Given that the height \( H \) is fixed at 10 m, we can substitute this into the volume equation:
\[ 10000 = L \times B \times 10 \]
Dividing both sides by 10 gives:
\[ L \times B = 1000 \quad (1) \]

### Step 3: Set Up the Surface Area Equation
The total inner surface area \( A \) of the walls of the hall is given as 1300 m². The formula for the surface area of the walls is:
\[ A = 2H(L + B) \]
Substituting \( H = 10 \):
\[ 1300 = 2 \times 10 \times (L + B) \]
This simplifies to:
\[ 1300 = 20(L + B) \]
Dividing both sides by 20 gives:
\[ L + B = 65 \quad (2) \]

### Step 4: Solve the System of Equations
Now we have two equations:
1. \( L \times B = 1000 \)
2. \( L + B = 65 \)

From equation (2), we can express \( B \) in terms of \( L \):
\[ B = 65 - L \]
Substituting this into equation (1):
\[ L \times (65 - L) = 1000 \]
Expanding this gives:
\[ 65L - L^2 = 1000 \]
Rearranging it leads to a quadratic equation:
\[ L^2 - 65L + 1000 = 0 \]

### Step 5: Solve the Quadratic Equation
To solve the quadratic equation \( L^2 - 65L + 1000 = 0 \), we can use the quadratic formula:
\[ L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -65, c = 1000 \):
\[ L = \frac{65 \pm \sqrt{(-65)^2 - 4 \times 1 \times 1000}}{2 \times 1} \]
Calculating the discriminant:
\[ (-65)^2 - 4000 = 4225 - 4000 = 225 \]
Thus,
\[ L = \frac{65 \pm 15}{2} \]
Calculating the two possible values for \( L \):
1. \( L = \frac{80}{2} = 40 \)
2. \( L = \frac{50}{2} = 25 \)

### Step 6: Find Corresponding Values for \( B \)
Using \( L + B = 65 \):
1. If \( L = 40 \), then \( B = 65 - 40 = 25 \).
2. If \( L = 25 \), then \( B = 65 - 25 = 40 \).

Thus, the dimensions of the hall are:
\[ L = 40 \, m, \quad B = 25 \, m \]

### Final Answer
The length and breadth of the hall are **40 meters and 25 meters**.

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