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Math Help - Optimization Help!

  1. #1
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    Optimization Help!

    The problem says...

    The trough in the figure is to be made to the dimension shown. Only the angle delta can be varied. What value of delta will maximize the trough's volume. (I attached a picture)

    Can someone please show me and explain to me how I would do a problem like this!
    Attached Thumbnails Attached Thumbnails Optimization Help!-90981018398581703cbfd29c2376528a.jpg  
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    Quote Originally Posted by KarlosK View Post
    The problem says...

    The trough in the figure is to be made to the dimension shown. Only the angle theta can be varied. What value of theta will maximize the trough's volume. (I attached a picture)

    Can someone please show me and explain to me how I would do a problem like this!
    Hi Karlos,

    The volume of the trough can be subdivided into the central cube
    and the two outside pyramids.

    To calculate volume, use

    V=(cross-sectional\ area)(length)

    hence you must express the trough's height "h" in terms of \theta

    cos\theta=\frac{h}{1}=h

    The cross-sectional area of the right and left triangles are

    \frac{1}{2}(h)(1)sin\theta=\frac{1}{2}cos\theta\ sin\theta

    Hence the volume of the trough is

    V=20cos\theta+20(2)\frac{1}{2}cos\theta\ sin\theta=20cos\theta(1+sin\theta)

    Theta giving max volume is found by differentiating this and setting it to zero, using the product rule of differentiation

    \frac{d}{d\theta}\left(20cos\theta(1+sin\theta)\ri  ght)

    20\left(cos\theta\ cos\theta+(1+sin\theta)(-sin\theta)\right)=0

    \left(cos^2\theta-sin\theta-sin^2\theta\right)=0

    1-sin^2\theta-sin\theta-sin^2\theta=0

    2sin^2\theta+sin\theta-1=0

    (2sin\theta-1)(sin\theta+1)=0

    For an acute angle use the first factor

    2sin\theta=1

    sin\theta=\frac{1}{2}

    \theta=sin^{-1}\frac{1}{2}=30^o
    Last edited by Archie Meade; April 28th 2010 at 05:18 AM. Reason: small typo
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  3. #3
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    Quote Originally Posted by KarlosK View Post
    The problem says...

    The trough in the figure is to be made to the dimension shown. Only the angle delta can be varied. What value of delta will maximize the trough's volume. (I attached a picture)

    Can someone please show me and explain to me how I would do a problem like this!
    basic trig w/ the two congruent right triangles on each side ...

    trapezoid height = \cos{\theta}

    top trapezoid base = 1 + 2\sin{\theta}

    cross-sectional area ...

    A = \frac{\cos{\theta}}{2}[(1+2\sin{\theta}) + 1]

    proceed to find the value of \theta that will maximize the cross-sectional area, and hence, the volume.
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    Quote Originally Posted by skeeter View Post
    basic trig w/ the two congruent right triangles on each side ...

    trapezoid height = \cos{\theta}

    top trapezoid base = 1 + 2\sin{\theta}

    cross-sectional area ...

    A = \frac{\cos{\theta}}{2}[(1+2\sin{\theta}) + 1]

    proceed to find the value of \theta that will maximize the cross-sectional area, and hence, the volume.
    Skeeter I have Area= cos(theta)+1/2sin2(theta) Is that incorrect? I attached how I came to that...
    Attached Thumbnails Attached Thumbnails Optimization Help!-gif.latex.gif  
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    Quote Originally Posted by Archie Meade View Post
    Hi Karlos,

    The volume of the trough can be subdivided into the central cube
    and the two outside pyramids.

    To calculate volume, use

    V=(cross-sectonal\ area)(length)

    hence you must express the trough's height "h" in terms of \theta

    cos\theta=\frac{h}{1}=h

    The cross-sectional area of the right and left triangles are

    \frac{1}{2}(h)(1)sin\theta=\frac{1}{2}cos\theta\ sin\theta

    Hence the volume of the trough is

    V=20cos\theta+20(2)\frac{1}{2}cos\theta\ \sin\theta=20cos\theta(1+sin\theta)

    Theta giving max volume is found by differentiating this and setting it to zero, using the product rule of differentiation

    \frac{d}{d\theta}\left[20cos\theta(1+sin\theta)\right]

    20[cos\theta\ cos\theta+(1+sin\theta)(-sin\theta)]=0

    20\left(cos^2\theta-sin\theta-sin^2\theta\right)=0

    1-sin^2\theta-sin\theta-sin^2\theta=0

    2sin^2\theta+sin\theta-1=0

    (2sin\theta-1)(sin\theta+1)=0

    For an acute angle use the first factor

    2sin\theta=1

    sin\theta=\frac{1}{2}

    \theta=sin^{-1}\frac{1}{2}=30^o

    How come you have sin^-1 instead of just sin at the end of this? Thanks for the help
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    A = \frac{\cos{\theta}}{2}[(1+2\sin{\theta}) + 1]

    A = \frac{\cos{\theta}}{2}[2+2\sin{\theta}]

    A = \cos{\theta}(1 + \sin{\theta})
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  7. #7
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    Quote Originally Posted by skeeter View Post
    A = \frac{\cos{\theta}}{2}[(1+2\sin{\theta}) + 1]

    A = \frac{\cos{\theta}}{2}[2+2\sin{\theta}]

    A = \cos{\theta}(1 + \sin{\theta})
    So after I have the area what would I do? Differentiate it? Can you tell me what steps I need to complete and then I can try to solve it?
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  8. #8
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    Quote Originally Posted by KarlosK View Post
    So after I have the area what would I do? Differentiate it? Can you tell me what steps I need to complete and then I can try to solve it?
    find \frac{dA}{d\theta} and determine the value of \theta that maximizes A.
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    Quote Originally Posted by KarlosK View Post
    How come you have sin^-1 instead of just sin at the end of this? Thanks for the help
    At the end, we have the sine of the angle..

    sin\theta=\frac{1}{2}

    the "inverse sine" retrieves the angle

    arcsin\left(\frac{1}{2}\right)=sin^{-1}\left(\frac{1}{2}\right)=30^o

    In your earlier post, you said you had

    cross-sectional area =cos\theta+\frac{1}{2}sin2\theta

    this is correct, as

    \frac{1}{2}sin2\theta=\frac{1}{2}2sin\theta\ cos\theta=sin\theta\ cos\theta

    Since the volume is 20 times the cross-sectional area,
    you only need to differentiate the cross-sectional area and set the result to zero.
    Then discover the value of theta that causes the derivative to be zero.

    \frac{d}{d\theta}\left[cos\theta(1+sin\theta)\right]=0

    Use the product rule as shown earlier

    or

    \frac{d}{d\theta}\left(cos\theta+\frac{1}{2}sin2\t  heta\right)=0
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