Question

A metal wire \(108\) meters long is bent to form a rectangle. If the area of the rectangle is maximum, then its dimensions are

• A. \( 28 \,\text{m}, 28 \,\text{m} \)
• B. \( 27 \,\text{m}, 27 \,\text{m} \)
• C. \( 25 \,\text{m}, 25 \,\text{m} \)
• D. \( 26 \,\text{m}, 26 \,\text{m} \)

Hint:

Among all rectangles with a fixed perimeter, the square encloses the maximum area.

Answer 1

The Correct Option is B

Step 1: Use the perimeter condition.
The total length of the wire is the perimeter of the rectangle: \[ 2(l + b) = 108 \Rightarrow l + b = 54 \] Step 2: Express area in one variable.
\[ A = l b = l(54 - l) \] Step 3: Maximize the area.
\[ \frac{dA}{dl} = 54 - 2l = 0 \Rightarrow l = 27 \] Step 4: Find breadth.
\[ b = 54 - 27 = 27 \] Step 5: Conclusion.
The rectangle of maximum area is a square of side \( 27 \) meters.