Question

A wire 64 m long is bent into the shape of a rectangle. What should be its dimensions so that the enclosed area is maximum?

• A. \(16 \text{ m}, 16 \text{ m}\)
• B. \(18 \text{ m}, 18 \text{ m}\)
• C. \(24 \text{ m}, 24 \text{ m}\)
• D. \(36 \text{ m}, 36 \text{ m}\)

Hint:

For a fixed perimeter: \[ \text{Maximum area rectangle} \Rightarrow \text{Square} \] So directly divide perimeter by 4: \[ \frac{64}{4}=16 \] Thus each side becomes \(16\) m.

Answer 1

The Correct Option is A

Concept: For optimization problems involving maximum area or minimum cost, we first express the required quantity in terms of a single variable and then use derivatives to find critical points. For a rectangle: \[ \text{Perimeter} = 2(l+w) \] and \[ \text{Area} = lw \] When perimeter is fixed, the rectangle having maximum area is always a square.

Step 1: Forming the perimeter equation.
Let the length of the rectangle be \(l\) metres and breadth be \(w\) metres. Given total wire length \(=64\) m. Therefore, \[ 2(l+w)=64 \] Dividing by 2: \[ l+w=32 \] Thus, \[ w=32-l \] Now the breadth has been expressed in terms of the length.

Step 2: Forming the area function.
Area of rectangle: \[ A=l\times w \] Substituting \(w=32-l\): \[ A=l(32-l) \] Expanding: \[ A=32l-l^2 \] Thus, the area function is: \[ A(l)=32l-l^2 \]

Step 3: Differentiating the area function.
Differentiate with respect to \(l\): \[ \frac{dA}{dl}=32-2l \] For maximum area: \[ \frac{dA}{dl}=0 \] Therefore, \[ 32-2l=0 \] \[ 2l=32 \] \[ l=16 \]

Step 4: Finding the breadth.
Using \[ w=32-l \] \[ w=32-16=16 \] Thus, \[ l=w=16 \] Hence the rectangle becomes a square.

Step 5: Verifying maximum condition using second derivative.
Differentiate again: \[ \frac{d^2A}{dl^2}=-2 \] Since \[ \frac{d^2A}{dl^2}<0 \] the area is maximum at \(l=16\). Hence the required dimensions are: \[ 16\text{ m} \times 16\text{ m} \]