Results 1 to 3 of 3

Math Help - Application of integration(the learning curve)

  1. #1
    Member
    Joined
    Sep 2011
    From
    Mumbai, Globally also known as 'Bombay', India
    Posts
    133
    Thanks
    2

    Post Application of integration(the learning curve)

    The production manager of an electronic company obtained the following function
    f(x)=1356.4x^{-0.3218}

    Where f(x) is the rate of labour hours required to assemble the x^{th} unit of a product. The function is based on the experience of assembling the first 50 units of the product. The company was asked to bid on a new order of 100 additional units.
    find the total labour hours required to assemble 100 units

    SOLUTION: -

    N= \int_{50}^{150}f(x)dx=\int_{50}^{150}1356.4x^{-0.3218}

    N= \left|\frac{1356.4x^{0.6782}}{0.6782}\right|_{50}^  {150}


    N= 2000\left|150^{0.6782}-50^{0.6782}\right|

    Using logarithm and anti-logarithm,we get,

    N= 2000[29.91-14.2]

    N=31420

    Hence company can bid estimating the total labour hours needed 31420
    Verify this answer,

    If I am wrong in finding the solution, reply me.
    Last edited by Vinod; December 18th 2011 at 04:34 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member ILikeSerena's Avatar
    Joined
    Dec 2011
    Posts
    733
    Thanks
    121

    Re: Application of integration(the learning curve)

    Hi Vinod!

    Your reasoning is correct.

    I do wonder though, whether you're supposed to make a continuity correction.
    This would mean adding 0.5 to both integral boundaries.
    Or is this out of scope of your material?

    Btw, in your problem statement your have 1356, while in your calculation you use 1356.4.
    This may be nitpicking, but are you supposed to do that?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2011
    From
    Mumbai, Globally also known as 'Bombay', India
    Posts
    133
    Thanks
    2

    Post Re: Application of integration(the learning curve)

    Quote Originally Posted by ILikeSerena View Post
    Hi Vinod!

    Your reasoning is correct.

    I do wonder though, whether you're supposed to make a continuity correction.
    This would mean adding 0.5 to both integral boundaries.
    Or is this out of scope of your material?

    Btw, in your problem statement your have 1356, while in your calculation you use 1356.4.
    This may be nitpicking, but are you supposed to do that?
    Hey,
    Thanks for your reply.I have inadvertently typed in the problem statement '1356'instead of '1356.4' I am sorry about it. I have rectified my error in the problem now.
    Last edited by Vinod; December 18th 2011 at 04:42 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Application of Integration
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 27th 2011, 07:55 AM
  2. Application of Integration-Mean
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 2nd 2009, 02:46 AM
  3. Help in Application of integration.
    Posted in the Calculus Forum
    Replies: 6
    Last Post: May 19th 2009, 07:04 AM
  4. application of integration
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 21st 2009, 03:23 PM
  5. integration of application
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 25th 2008, 12:50 PM

Search Tags


/mathhelpforum @mathhelpforum