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My Maths Project. Kinda Confuse and Lost.

  1. Aug 9, 2006 #1


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    http://img74.imageshack.us/img74/3570/pict00011ms5.jpg [Broken]

    The picture shows a gutter which is a trapezium same leg. Given the length of the leg are 10 cm including the base of the trapezium. Count the part of the open top part of the trapezium , d , so that the gutter can support the maximum volume of water. Give the reason to support your working answer.

    My Working
    Im also not really sure of the question they asking. Sorry cause i had to translate my maths from malay to english. If im not wrong, i must count the length of the open top part. Then count the volume of maximum water that can be support by the gutter.

    I found the length of " d " not the whole top part. Assume the whole gutter was a heksagon. So i use the formula ( n - 2 ) X 180 degree. I get the whole inside angleis 720 then divide by 6. So each angle is 120 degree. Then i cut the angle to half then use sin to count 1 part of the length of " d ". Add up the total i got the length of " d " is 20 cm.

    Now im stuck here because im not sure how i want to count next. I notice that i need to use Differentiation to count the volume of water. But im not sure how. This my working.

    Please correct me if Im wrong. But im sure that i did some mistake. IF i did not provide enough information please tell. Sorry for my bad language cause i kinda learn my science and maths subject in Malay. Really sorry And Thx
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Aug 9, 2006 #2
    what math is this? Calculus? have you learned lagrangian multipliers...perhaps you dont' need this.

    anyways you CANNOT assume the shape forms a regular hexagon because the question is what is the maximum volume attained by the 3 sides of 10cm. SO you must find the volume of this so called Trapezium without solving for d...Assuming it is a piped(parallel) 3D shape..this becomes a problem of finding the MAX area on the 2D area of 3 sides + d.
    ONce you have found a formula for the area of the trapezoid 3 sides + d.
    You will do some calculus. to find d...so remember find the volume before finding d.
  4. Aug 9, 2006 #3


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    You really want to maximize the cross section area. Then for any length, l, there will be the maximum volume. To find that, you will need to find the "height": A= h(b1+ b2)/2= h(d+ 10)/2.

    You have a trapezoid with base of length d, another, shorter, base of length 10 and two legs of length 10. Okay, drop perpendiculars from the shorter base to the longer. If the shorter base has length 10, then the two sections on either side of that, on the longer base is d- 10. Each of the sections has length (d-10)/2. So you you on the ends you have two right triangles. You know the hypotenuse is of length 10 and one leg has length (d-10)/2. Use the Pythagorean theorem to find the length of the other leg, the height of the parallelogram in terms of d. Find the value of d that maximizes that.

    Since this is posted under "precalculus", the formula for area will be a quadratic in d and you can maximize that by completing the square.
  5. Aug 9, 2006 #4


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    Ok i found the correct D this time i hope so. Sorry i kinda ask my fren to scan the pic. Both legs are 10 cm. Im now having problem finding the height. Can you explain this formula to me

    "height": A= h(b1+ b2)/2= h(d+ 10)/2.

    Sorry im last year in secondary school.
  6. Aug 9, 2006 #5


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    You've "separated" that in the wrong place! I was saying that the formula for area of a trapezoid is A= h(b1+ b2)/2 where b1 and b2 are the two bases and h is the height (measured perpendicular to both bases). Then I said that you can find the height by dividing the trapezoid into a rectangle and two right triangles at the two ends. Since the longer base has length d and the shorter length 10, the "rectangle" will have length 10 which leaves d-10 for the bases of the two right triangles together. The length of the base of one right triangle is (d-10)/2 and the hypotenuse is 10. The other leg is h and, by the Pythagorean theorem, h2+ {(d-10)/2}2= 100. Therefore,
    [tex]h= \sqrt{100- \frac{(d-10)^2}{4}}= \sqrt{100- \frac{d^2}{4}+ 5d- 25}[/tex]
    [tex]h= \sqrt{75+ 5d- \frac{d^2}{4}}[/tex]
    Use that to find the area in terms of d and determine which d maximizes that.
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