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# 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 $$= 4p$$
$$4 \times \frac{p}{4} = p.$$

## Discussion (3)

• Ismail Rokeeb

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

Reply
• 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 $$= 4p$$
$$4 \times \frac{p}{4} = p.$$

Reply