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Optimization Problem

  1. Jul 21, 2012 #1
    1. The problem statement, all variables and given/known data
    If the area of the window is 3m² what are the dimensions of a rectangle and (isosceles)triangle that will minimize the perimeter.
    http://s12.postimage.org/r1czis8jh/optimize.png --DIAGRAM

    2. Relevant equations

    Let AreaTotal=3
    Let Areatotal=Atriangle+Arectangle

    Arectangle=lxw
    Atriangle =bxh/2

    3. The attempt at a solution

    so because the triangle is isosceles the base and height are = x.

    so Atriangle = x*x/2=x²
    Arectangle= l*w=xy

    Atotal = Atriangle+Arectangle
    3=x²/2 + xy
    y=(6-x²)/2x(1)-- y in terms of x

    Prectangle= 2y+x
    = 2(6-x²)/2x +x
    =(6-x²)/x +x
    =(6-x²+x²)/x
    =6/x
    it is only 2y+x because the perimieter of the window does not include the 4 rectangle side.

    Ptriangle= x+x+√(2x²)
    =2x+√(2x²)
    The perimeter of the triangle includes opposite adjacent and hypotanuse sides.

    Perimeter Total=Ptriangle + Prectangle.
    Ptotal=6/x + 2x+√(2x²)

    Ptotal=¸(6+2x²+x²*√(2))/x


    Domain: XER, X Cannot Be Zero.

    Now before i go any further, i was wondering if i could get some help ion figuring out if the equation i derived would be correct... Because from here, i would Take the derivative and determine critical points and a minimum value and then solves for the sides.. Anyway i appreciate any help :D
     
    Last edited: Jul 21, 2012
  2. jcsd
  3. Jul 21, 2012 #2

    LCKurtz

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    But you don't include the whole perimeter of the triangle in the exterior perimeter either so the 2 highlighted below should be 1.
    And writing ##x\sqrt 2## as ##\sqrt{2x^2}## will only complicate the derivative later.
     
  4. Jul 21, 2012 #3
    Ah yea you`re right, i dont know why i was including the side between the rectangle and triangle.. And okay, i`ll keep it as √(2x²), thank you for the tip.

    i`m gonna rewrite it again, do you mind checking again?
     
  5. Jul 21, 2012 #4
    So...

    Ptriangle = x+2√x²
    Prectangle= 2y+x=6/x

    Ptotal = ptriangle+prectangle
    ptotal=6/x + x + √2x²
    ptotal=( 6 + x² + x√2x²)/x

    p`total=( x(2x+√2x²-2x/√(2x²))-(6+x²+ x√2x²) ) / x²

    p`total=(2x² + x√2x²-((x²/√(2x²)) - 6 - x² - x√2x²)/x²
    p`total= (x² - x²/√(2x²) - 6)/x²
    p`total= (x²(√(2x²)) - 6(√(2x²)) -x²)/x²(√(2x²))
    p`total= x(x²√2 -6√2 - x)/x²(√(2x²))
    p`total=(x²√2 -6√2 - x)/x(√(2x²))

    now i have a quadratic in the numerator..
    p`=0
    0=x²√2 -x -6√2

    Is this correct ?
     
  6. Jul 21, 2012 #5

    LCKurtz

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    One leg plus hypotenuse gives ##x+x\sqrt 2##.
     
  7. Jul 21, 2012 #6
    ptriangle= x+x√2
    prectangle=x+2y
    =2(6-x²)/2x+x
    =6/x

    Ptotal=ptriangle+prectangle
    Ptotal= 6/x + x +x√2

    ptotal= (6 + x² + x²√2)/x

    p`total= ( (0 +2x + 2x√2)(x) - (6 + x² + x²√2)(1) )/x²

    p`total= (2x² + 2x²√2 -6 - x² - x²√2)/x²
    p`total=(x² + x²√2 -6)/x²

    How about this :O
     
  8. Jul 21, 2012 #7

    LCKurtz

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    That's better. So what do you get for the dimensions and the minimum perimeter?
     
  9. Jul 21, 2012 #8
    okay so..
    if p`total=(x² + x²√2-6)/x²

    p`total= dne
    x=0 (restricted by domain)

    p`total=0
    0=x² + x²√2-6
    6 = x²(1+√2)
    x=√(6/(1+√2))
    x= +/-1.576

    I so plotted this on a line..

    **i made a chart here but it didnt work.. i basically took the values of the slope before and after the point 1.58

    so what i found was

    + -1.58 -
    positive before, negative after, so therefore that was a maxiumum.
    and then

    - 1.58 +
    negative before positive after, so i found that this point was a minimum, so it had to be the minimum value.
    **i plugged in values before and after and determined the sign of the slope.**

    i found that 1.58 was a minimum value so therefore it is a CP.

    So if this is the minimum value.

    y=(6-x²)/2x
    plug inx=1.58
    y=1.109
    y=1.11

    Hypotenuse=√2x² = 2.23

    so therefore dimensions are

    x=1.58 for the triangle and rectangle, y=1.11 for the rectangle,
    and hypotenuse of the triangle is hyp= 2.23
     
  10. Jul 21, 2012 #9

    LCKurtz

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    Your numbers are correct, but you keep doing that. You have written ##\sqrt 2x^2## which is the same as ##x^2\sqrt 2## which is not what you mean. If you insist on calling ##x\sqrt 2## as ##\sqrt{2x^2}## you need parentheses around the ##2x^2## or the top bar on the square root sign. It's easier to not put the ##x## under the square root sign in the first place, and in this situation it is best to write the ##x## before the ##\sqrt 2## to avoid such misunderstandings.
     
  11. Jul 21, 2012 #10
    i`m really sorry about that that you for reminding me again, i need to watch that or people aren`t gonna understand me :(. Thank you very much for your help i really appreciate it :D.
     
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