A wire of length 36meters is bent in the form of a rectangle. Find its dimensions if the area of the rectangle is maximum

Answer

To find the dimensions of a rectangle formed by a wire of length 36 meters that maximizes the area, we can follow these steps:

### Step 1: Define the Variables
Let \( L \) be the length of the rectangle and \( B \) be the width of the rectangle.

### Step 2: Write the Perimeter Equation
Since the wire is bent into a rectangle, the perimeter \( P \) can be expressed as:
\[ P = 2L + 2B \]
Given that the length of the wire is 36 meters, we have:
\[ 2L + 2B = 36 \]
Dividing the entire equation by 2 gives:
\[ L + B = 18 \]

### Step 3: Express Area in Terms of One Variable
The area \( A \) of the rectangle is given by:
\[ A = L \times B \]
From the perimeter equation, we can express \( B \) in terms of \( L \):
\[ B = 18 - L \]
Substituting this into the area formula gives:
\[ A = L(18 - L) = 18L - L^2 \]

### Step 4: Differentiate the Area Function
To find the maximum area, we differentiate \( A \) with respect to \( L \):
\[ \frac{dA}{dL} = 18 - 2L \]

### Step 5: Set the Derivative to Zero
To find the critical points, set the derivative equal to zero:
\[ 18 - 2L = 0 \]
Solving for \( L \):
\[ 2L = 18 \implies L = 9 \]

### Step 6: Find the Width
Now, substitute \( L = 9 \) back into the equation for \( B \):
\[ B = 18 - L = 18 - 9 = 9 \]

### Step 7: State the Dimensions
Thus, the dimensions of the rectangle that maximize the area are:
\[ L = 9 \text{ meters}, \quad B = 9 \text{ meters} \]

### Conclusion
The rectangle with maximum area formed by the wire of length 36 meters has dimensions of 9 meters by 9 meters.

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