**Find the dimensions of a rectangle of greatest area which has a fixed perimeter ...**

### Question

Find the dimensions of a rectangle of greatest area which has a fixed perimeter \(p.\)

### Options

A) square of sides \(p\)

B) square of sides \(2p\)

C) square of sides \(\frac{p}{2}\)

D) square of sides \(\frac{p}{4}\)

The correct answer is D.

### Explanation:

Let the rectangle be a square of sides \(\frac{p}{4}.\)

So that perimeter of square \(= 4p\)

\(4 \times \frac{p}{4} = p.\)

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## Discussion (3)

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In my opinion, a rectangular shape consist of a length( long size ) and a breadth ( of short length)

The questions says a dimensions of greatest area ( i.e the length)

Then,

If perimeter p,= 2(l+b)

The greatest area L should be = (p/2) - b

Perimeter of rectangle=P

Perimeter=2(l+b)

2(l+b)=P

P/2=l+b

Let's take l and b as the same since it is a square that is to be solved

P/2=2l

P/4=l

So the length of the sqaure is p/4

Let the rectangle be a square of sides \(\frac{p}{4}.\)

So that perimeter of square \(= 4p\)

\(4 \times \frac{p}{4} = p.\)