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


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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

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

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