1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Maximize Area

  1. May 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Note the picture

    2. Relevant equations

    3. The attempt at a solution

    A of both trianles = xsinQ[tex]\sqrt{x^2 - x^2(sinQ})^2[/tex]

    A of rectangle = (w-2x)(xsinQ)

    Total A = (w-2x)(xsinQ) + xsinQ[tex]\sqrt{x^2 - x^2(sinQ)^2}[/tex]

    A = wxsinQ + 2x2sinQ + x2sinQcosQ

    Ax = wsinQ + 4xsinQ + 2xsinQcosQ = 0

    -wsinQ = x(4sinQ + 2sinQcosQ)

    x = -w/(4 + 2cosQ)

    AQ = wxcosQ + 2x2cosQ + x2(sinQ)2 - x2(cosQ)2 = 0

    It gets pretty messy when i plug in x, are there anny errors so far

    Attached Files:

    • 5.pdf
      File size:
      9.4 KB
  2. jcsd
  3. May 17, 2009 #2
    anybody, can i get some help
  4. May 17, 2009 #3


    Staff: Mentor

    What's the problem you are trying to solve? The picture doesn't give me any idea of what you're trying to do.
  5. May 17, 2009 #4
    you must Maximize the area, note the title
  6. May 18, 2009 #5


    Staff: Mentor

    Some context would have been nice, such as that the area is the cross-sectional area of what looks to be a trough, formed from sheet metal that is W units wide.

    Your picture doesn't convey that information, which would have been helpful.

    You have a sign error in this line:
    The first '+' sign should be '-'. That makes Ax = WxsinQ - 4xsinQ + 2xsinQcosQ = sinQ[W -4x + 2xcosQ].

    The sign error also affects AQ, which I get as xWcosQ -2x2cosQ + x2cos2Q - x2sin2Q.
    About the only thing I can think of to do is to factor x out of each term in this partial.

    The critical points are where Q = 0 or when x = 0, either of which gives you an area of 0, and whatever you get from some messy equations that remain. One of these is W - 4x + 2xcosQ = 0, which you can solve for x in terms of Q (and W), or solve for Q in terms of x (and W). Whichever variable you solve for, substitute that in the other equation that comes from AQ.
  7. May 18, 2009 #6
    Ya, this is pretty much where i ran into problems, i pluged in my x = w/(4-2cosQ) into A sub Q

    AQ = 0 = (w2cos)/(4-2cos) - 2w2cos/(4-2cos)2 + (2w2cos2)/(4-2cos)2 - (w2sin2)/(4-2cos2)

    I would have to solve that for Q (i got lazy and didnt type them, just assume they are there), which means (tell me if im wrong), i only have to solve the numerators of the expression for 0 in terms of w

    I got Q = 0 & 2pi which would make the gutter flat, what happened
  8. May 18, 2009 #7


    Staff: Mentor

    I didn't work it through that far, so I'll assume that the work you have in post 6 is correct. You can't just "solve the numerators," as you put it, since not all of your expressions have the same denominator; the first has a denominator of 4 - 2cosQ. You'll need to multiply that term by (4 - 2cosQ)/(4 - 2cosQ), and then you can factor out 1/(4 - 2cosQ)2 from all four terms. Finally, you'll need to see what values of Q (theta) make AQ equal to 0.

    As I said before, Q = 0 and x = 0 are critical values, but they produce a trough with cross-sectional area 0, so they are not the values you want.
  9. May 18, 2009 #8
    ok i was able to factor out the w^{2} and the (4-2cos)^{2} and I got

    4cos - 2cos^{2} - 2cos + cos^{2} - sin^{2} = 2cos - con^{2} - sin^{2} = 0

    I got Q = pi/3, 5pi/3, 7pi/3

    how do i calc which one is the maximum, there are many possibilities
  10. May 18, 2009 #9


    Staff: Mentor

    Due to the fact that you're working with a physical object, a trough, there is a natural range of values for Q (theta). That should help you eliminate many of the possibilities. If Axx < 0 and AQQ < 0, you're at a local maximum.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook