Oxford PAT 2010, Questions 5 and 6

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10 thoughts on “Oxford PAT 2010, Questions 5 and 6

  1. For question 6, what I ve done is said that the largest possible area you can get is a square. But since you cannot do a square I’ve labeled the width and length with W and l respectively. So if l = w then l and W = L/4. So how the rectangle is changing would be given by the function : (l + n) . (W -n) . *we are assuming that l>w
    So lets substitute l and W= L/4. Then (L/4 + n) . (L/4 – n). If we expand it we are left with
    L^2/16 – n^2. So the largest possible area of the rectangle would be given by the expression
    L^2 / 16 -n^2 . As I can see from your answer above the answer is just L^2/16. Can you tell me if this makes any sence to you? And is it correct?

    1. Yes your answer does make sense, and is actually a really nice idea, but there are a couple of problems with the way you have described it.

      First, you say “you cannot do a square”, but a square is allowed (because a square is a rectangle), and in fact the maximum area is when the rectangle is a square.

      Second, with your calculations you have actually got the answer right in quite a nice elegant way, but you haven’t taken the final step and finished it off: your answer points out that in general the sides of the rectangle will be (L/4 + n) and (L/4 – n), and therefore the area is L²/16 – n²; since n² must be greater than or equal to zero, the area is at a maximum when n = 0, and that area is L²/16.

  2. For question 6, if the rectangle is a square then the area is (L/4)*(L/4) = (L^2/16). Therefore the area within the rope rectangle must always be (L^2/16). It’s like a necklace – no matter what shape you make by pulling the edges of the necklace, the area contained within the edges will always be the same.

    1. The first sentence you’ve written is true, but the others are not. Try taking an actual necklace, putting one finger from each hand in the necklace, and then pulling your hands apart as far as they will go; you’ll see that the area enclosed by the necklace is almost zero. The maximal area enclosed is when the necklace encloses a square, but the question asks you to derive that, rather than just assert it.

      1. I feel like an idiot. Thank you for clearing up that terrible misconception, and thank you for this excellent resource.

        1. It’s a pleasure. And don’t worry — we all have misconceptions to clear up from time to time. One idea I’ve found useful in the past is, if you find yourself thinking of a principle in words, try to prove it; if it’s wrong you won’t be able to prove it and you’ll probably find a counterexample.

    1. Well actually the segment I have drawn is the one that contains the vertical line hatching, and this is 2/3 the area of the circle. But the other segment is 1/3 the area.

      You can tell this because we can see that the angle I have labelled in the diagram is π/3 (because it is arcsin(0.5R / R)). By symmetry its mirror image in the x-axis is also π/3, and therefore the sum of the two is 2π/3; since the total ange is 2π this means the segment area is ((2π/3)/2π) tims the area of the circle, which is 1/3 the area of the circle.

    1. Hi Kara, thanks for your question.

      It’s a good question, and actually there is nothing special about the method I have chosen. I could just as well have said that one side is x, and then the adjacent side must be L/2-x. Then when you work out the area and differentiate it you get x = L/4 at the maximum, and its adjacent side is L/4. So if anything this would be neater…

      I think the necessary thought process for answering this question in a reasonably rigorous way is ‘Find any equation for the area in terms of some unknown and L; solve for the unknown at the maximum to find the value of the unknown in terms of L; plug in the solved value to find the area in terms of L’. But there are absolutely no rules about how to express the line lengths to generate the initial equation: anything that works is good.

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