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Quite a long 'bounds' question

  1. Feb 17, 2007 #1
    hope you can help with this question, its a gcse paper (16 year olds) and I think i may have either made a mistake, been doing it the wrong way, or simply found the answer but me being paranoid thinks the working is too long and answer not simple enough

    thnx for the help

    1. The problem statement, all variables and given/known data

    The length of a side of the square base of a pyramid is x metres.
    The vertical height is y metres.

    Both measurements are correct to the nearest metre.

    question) Find an expression for the difference between the upper bound and the lower bound of the volume of the pyramid.
    Give your answer in its simplest form.

    2. Relevant equations

    volume of pyramid = (1/3)base x height

    3. The attempt at a solution

    my notation: (lb) means lower bound of, so (lb)y would mean the lower bound of y, and so on.

    p.s. I know some (well alot) are redundant but I stuck them in to make the workings easier to read. Also, i should really learn Latex but I havnt got round to yet so please try to read my non-latex stuck lol.

    (lb)x = x - (1/2) x (1/10)x
    (lb)x = x - (1/20)x
    (lb)x = (19/20)x

    therefore: (ub)x = (21/20)x
    (lb)y = (19/20)y
    (ub)y = (21/20)y

    (lb)base = (lb)x^2
    (lb)base = (19x/20)^2
    therefore: (ub)base = (21x/20)^2

    (lb)volume = (1/3) x (lb)b x (lb)h
    v = (1/3) x (19x/20)^2 x (19y/20)

    (ub)volume = (1/3) x (21x/20)^2 x (21y/20)

    difference = (ub)vol - (lb)vol
    d = (1/3) x (21x/20)^2 x (21y/20) - (1/3) x (19x/20)^2 x (19y/20)
    d = (21y x (21x/20)^2)/60 - (19y x (19x/20)^2) /60
    d = (21y x (21x/20)^2 - 19y x (19x/20)^2) /60

    is this the simplest form? seem rather complex to me.

    I tried factorising the brackets and doing all sorts but I couldn't get it any cleaner than this.

    thnx for your help
  2. jcsd
  3. Feb 17, 2007 #2


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    Science Advisor

    PLEASE, PLEASE, PLEASE do not use "x" to represent the length of the base AND as a "times" sign. If you must use a times sign (rather than just "xy") use *. Also use ( ). I THINK you mean (x- 1/2) (1/10)x.

    Now, the problem asked about the volume of a pyramid which you say is given by V= (1/3)base * height or, here V= (1/3)xy. Actually, that's impossible- it has the wrong units. If x and y are measured in meters, then xy has units of "square meters", not "cubic meters" which it would have to be in order to give volume. Are you sure it's not (1/3) base2*height?
    Are you saying here that the largest x can be is (x+ 1/2)(1/10)x? Where is that 1/10 from? If a length, x, is measured "to the nearest meter", the something as high as 6.4999999 meters would be given as 6 meters. "x meters, to the nearest meter" could be as large as x+ 1/2 meters.

    Same for y: the largest it could be (upperbound) is y+ 1/2, smallest (lowerbound) is y- 1/2.

    No only does it seem complex, it makes no sense to me!
    The upperbound for the volume is (1/3)(x+1/2)2(y+ 1/2) and the lowerbound is (1/3)(x-1/2)2(y-1/2).

  4. Feb 17, 2007 #3
    o rite thnx, i think i no where i might have gone wrong then, will go back and try again, thnx
  5. Feb 18, 2007 #4
    'base' as used in this formula is shorthand for 'area of the base' and so is x^2
  6. Feb 27, 2007 #5
    Hi I got everything up to the last part and figured out that

    d(difference between ub and lb) = 1/3(x+1/2)^2(y+1/2) - 1/3(x-1/2)^2(y-1/2)

    but I'm not quite sure how to simplify it.

    Would it require me to expand the brackets then factorise? (although that seems strangely long for a 3 mark question)

    Would it involve moving parts of the equation across the equal sign?

    or is it incredibly simple and I'm just missing something badly...

    Thanks, any help would be greatly appreciated.
  7. Feb 27, 2007 #6
    any ideas, anyone?

    edit: is there any chance that this is the answer?
    Last edited: Feb 27, 2007
  8. Feb 28, 2007 #7
    Yes, exactly as shown by HallsofIvy above
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