Results 1 to 6 of 6

Math Help - Minimization problem

  1. #1
    Newbie
    Joined
    Dec 2011
    Posts
    8

    Unhappy Minimization problem (multiple choice(

    A cylindrical can is to contain 2000 cu. inches of liquid. What dimensions will minimize the amount of metal used in the construction of the can?
    Choices:
    a. D = 6.828in, H = 13.656
    b. D = 6.424 in, H = 13.555
    c. D = 6.222 in , H = 13.757
    d. D = 6.626 in, H = 13.454

    I honestly dont know where to start
    Last edited by MathMinors; December 8th 2011 at 02:20 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5

    Re: Minimization problem (multiple choice(

    Setting r the radious of the base of the can and h its height, then we have...

    v= \pi\ r^{2}\ h (1)

    a= \pi\ r^{2} + 2 \pi r h (2)

    From (1) You derive...

    h= \frac{v}{\pi r^{2}} (3)

    ... so that from (2) and (3) You obtain...

    a= \pi r^{2} + \frac{2 v}{r} (4)

    Because v is known, Your task is to find the value of r that minimizes a...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2011
    Posts
    8

    Re: Minimization problem (multiple choice(

    Quote Originally Posted by chisigma View Post

    a= \pi\ r^{2} + 2 \pi r h (2)
    circular area of bot and top included ?
    a= 2\pi\ r^{2} + 2 \pi r h

    thnks btw
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5

    Re: Minimization problem (multiple choice(

    The cilinder has been considered 'open on the top'... if You want to include the top circular area the procedure is similar and the task is left to You...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5

    Re: Minimization problem (multiple choice(

    Taking into account the top and bottom circular areas the equations become...

    v=\pi\ r^{2}\ h (1)

    a=2\ \pi\ r\ (r+h) (2)

    ... and from them...

    a= 2\ \pi\ r^{2} + \frac{2\ v}{r} (3)

    Now applying the stand procedure we find the r for minimal area from the equation...

    \frac{d a}{dr}+ 4\ \pi\ r - \frac{2\ v}{r^{2}}=0 (4)

    ... the solution of which is...

    r=(\frac{v}{2\ pi})^{\frac{1}{3}} (5)

    For curiosity I mesured the dimensions of a can of 'Red Kidney-Beans' [my preferred ...] finding r=3.75,\ h=11.5\ \implies v=508. Applying (5) for v=508 we obtain an optimum value r=4.32... not very far from r=3.75 which requires however little more metal ...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Sep 2007
    Posts
    46
    Thanks
    1

    Re: Minimization problem

    Dont forget that not only are they trying to minimize the metal used in the can, but also trying to optimize shipping and storage as well as production. Perhaps their cans fit better in boxes or something?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Minimization Problem.
    Posted in the Calculus Forum
    Replies: 9
    Last Post: November 11th 2010, 07:48 PM
  2. Minimization problem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 22nd 2009, 03:03 PM
  3. Minimization problem - please help!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 14th 2009, 06:54 PM
  4. Minimization Problem
    Posted in the Math Challenge Problems Forum
    Replies: 2
    Last Post: August 17th 2007, 12:54 AM
  5. minimization problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 13th 2005, 04:16 AM

Search Tags


/mathhelpforum @mathhelpforum