Results 1 to 8 of 8

Math Help - Percentage increase in area of rectangle given increase in perimeter

  1. #1
    Junior Member
    Joined
    Aug 2010
    Posts
    31

    Percentage increase in area of rectangle given increase in perimeter

    One side of a rectangle is three times the other. If the perimeter increases by 2%, what is the percentage increase in area?

    I've started with these few lines:
    Perimeter, P = 2x + 2x + x + x = 8x
    Area, 3x^2 = 3(p^2)/64
    delta P= 2%
    delta A = dA/dP . delta P
    delta A = 3/64 (2p). 2%

    The answer is 4% but it doesn't look like delta A will be 4%!

    Any clues where I went wrong? Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,234
    Thanks
    27
    Quote Originally Posted by darkvelvet View Post
    One side of a rectangle is three times the other.
    P = 3x+3x +x+x
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Educated's Avatar
    Joined
    Aug 2010
    From
    New Zealand
    Posts
    399
    Thanks
    10
    Quote Originally Posted by darkvelvet View Post
    P = 2x + 2x + x + x = 8x
    Any clues where I went wrong? Thanks.
    How does 2+2+1+1 = 8?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Aug 2010
    Posts
    31
    Sorry, it should be like this:

    P = 3x + 3x + x + x = 8x
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570

    Area and length of similar shapes

    Hello darkvelvet
    Quote Originally Posted by darkvelvet View Post
    One side of a rectangle is three times the other. If the perimeter increases by 2%, what is the percentage increase in area?

    I've started with these few lines:
    Perimeter, P = 2x + 2x + x + x = 8x
    Area, 3x^2 = 3(p^2)/64
    delta P= 2%
    delta A = dA/dP . delta P
    delta A = 3/64 (2p). 2%

    The answer is 4% but it doesn't look like delta A will be 4%!

    Any clues where I went wrong? Thanks.
    First, a few basics:
    • The percentage increase in a quantity is the actual increase divided by the original amount, multiplied by 100. So you should have \dfrac{\delta P}{P}\times100=2, or \dfrac{\delta P}{P}=2%, not \delta P = 2%.
      .
    • Since the rectangle always has the same shape, its area is proportional to the square of the length of one of its sides, and therefore is proportional to the square of its perimeter. So A = kP^2.


    Differentiating this we get:
    \dfrac{dA}{dP}=2kP\approx\dfrac{\delta A}{\delta P}

    \Rightarrow \dfrac{\delta A}{A} \approx\dfrac{2kP\delta P}{A}\approx \dfrac{2kP\delta P}{kP^2}}
    \approx2\dfrac{\delta P}{P}

    So if \dfrac{\delta A}{A}=2%, then \dfrac{\delta P}{P}}\approx 4%.

    In fact, it's even easier without calculus. Note that, if the lengths of corresponding sides of two similar shapes are in the ratio 1:n, then their areas are in the ratio 1:n^2.

    A 2% increase in the length means an increase in the ratio 1:1.02; so the area increases in the ratio 1:1.02^2 = 1 : 1.0404, or 4.04%.

    This has the added advantage of being an exact answer, whereas the calculus method gives only an approximation.

    Grandad
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Aug 2010
    Posts
    31
    I get all the workings except for this:
    A = kP^2.

    What is k?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Hello darkvelvet
    Quote Originally Posted by darkvelvet View Post
    I get all the workings except for this:
    A = kP^2.

    What is k?
    It is the 'constant of proportionality'. If \displaystyle A is proportional to P^2, then we can write
    A = kP^2, where \displaystyle k is a constant.

    Using your original method, you (correctly) got
    A = \dfrac{3P^2}{64}
    So
    k = \frac{3}{64}

    As you'll see from my working, \displaystyle k is eliminated in the subsequent equations, so you didn't really need to evaluate it.

    Grandad
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Aug 2010
    Posts
    31
    Yay! Now I finally understand everything. Thanks so much!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Percentage increase
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 7th 2011, 12:19 AM
  2. Percentage increase.
    Posted in the Algebra Forum
    Replies: 1
    Last Post: August 3rd 2010, 03:49 AM
  3. percentage increase
    Posted in the Algebra Forum
    Replies: 4
    Last Post: February 25th 2009, 04:31 AM
  4. Increase Percentage
    Posted in the Algebra Forum
    Replies: 7
    Last Post: April 24th 2008, 02:02 PM
  5. Percentage Of Population Increase
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 30th 2007, 07:38 AM

/mathhelpforum @mathhelpforum