Results 1 to 10 of 10

Math Help - optimization problems

  1. #1
    Newbie
    Joined
    Mar 2007
    Posts
    8

    optimization problems

    1) a rectangular backyard playpen for child is to be enclosed with 16 m of fliexible fencing. what dimensions of the rectangle will provide the maximum area for the child to play?

    2) a rectangular corral is to be enclosed along the side of a horse barn with the barn serving as one side of the corral. what dimensions of the corral using 40 m of fencing, will enclose the maximum area for the horses?

    3) a rectangular garden plot requires an area of 32 m squared ...for the variety of vegetables that are tobe planted. what dimensions of the plot will use the lead amount of fencing to enclose the garden?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by juiicycouture View Post
    1) a rectangular backyard playpen for child is to be enclosed with 16 m of fliexible fencing. what dimensions of the rectangle will provide the maximum area for the child to play?

    2) a rectangular corral is to be enclosed along the side of a horse barn with the barn serving as one side of the corral. what dimensions of the corral using 40 m of fencing, will enclose the maximum area for the horses?

    3) a rectangular garden plot requires an area of 32 m squared ...for the variety of vegetables that are tobe planted. what dimensions of the plot will use the lead amount of fencing to enclose the garden?
    ALWAYS START OPTIMIZATION PROBLEMS BY DRAWING DIAGRAMS! see below, i drew some for you.

    1) a rectangular backyard playpen for child is to be enclosed with 16 m of fliexible fencing. what dimensions of the rectangle will provide the maximum area for the child to play?
    2x + 2y = 16 = P ..........P is for perimeter

    we want Area = A = xy to be a maximum

    since 2x + 2y = 16
    => x = 8 - y

    so A = (8 - y)y = 8y - y^2

    for max, we want A' = 0
    => A' = 8 - 2y = 0
    => y = 4

    But x = 8 - y
    => x = 4



    2) a rectangular corral is to be enclosed along the side of a horse barn with the barn serving as one side of the corral. what dimensions of the corral using 40 m of fencing, will enclose the maximum area for the horses?
    this is very similar to the last one

    x + 2y = 40

    want A = xy to be max

    since x + 2y = 40
    => x = 40 - 2y

    => A = (40 - 2y)y = 40y - 2y^2

    for max, A' = 0
    => A' = 40 - 4y = 0
    => y = 10

    but x = 40 - 2y
    => x = 20



    3) a rectangular garden plot requires an area of 32 m squared ...for the variety of vegetables that are tobe planted. what dimensions of the plot will use the lead amount of fencing to enclose the garden?
    A = xy = 32

    we want Perimeter = P = 2x + 2y to be minimum

    since xy=32
    => x = 32/y

    => 2(32/y) + 2y = P
    => P = 64/y + 2y = 64y^-1 + 2y

    for min, P' = 0
    => P' = -64y^-2 + 2 = 0
    => -64 + 2y^2 = 0
    => 2y^2 = 64
    => y^2 = 32
    => y = +/- sqrt(32)
    so y = sqrt(32) since we can't have negative area

    but x = 32/y
    => x = 32/sqrt(32)
    => x = sqrt(32)
    Attached Thumbnails Attached Thumbnails optimization problems-question-1.gif   optimization problems-question-2.gif   optimization problems-question-3.gif  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by juiicycouture View Post
    1) a rectangular backyard playpen for child is to be enclosed with 16 m of fliexible fencing. what dimensions of the rectangle will provide the maximum area for the child to play?

    2) a rectangular corral is to be enclosed along the side of a horse barn with the barn serving as one side of the corral. what dimensions of the corral using 40 m of fencing, will enclose the maximum area for the horses?

    3) a rectangular garden plot requires an area of 32 m squared ...for the variety of vegetables that are tobe planted. what dimensions of the plot will use the lead amount of fencing to enclose the garden?
    hold on, you're in pre-calculus?

    that means you don't know how to find a derivative right?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by juiicycouture View Post
    1) a rectangular backyard playpen for child is to be enclosed with 16 m of fliexible fencing. what dimensions of the rectangle will provide the maximum area for the child to play?

    2) a rectangular corral is to be enclosed along the side of a horse barn with the barn serving as one side of the corral. what dimensions of the corral using 40 m of fencing, will enclose the maximum area for the horses?

    3) a rectangular garden plot requires an area of 32 m squared ...for the variety of vegetables that are tobe planted. what dimensions of the plot will use the lead amount of fencing to enclose the garden?
    juiicycouture, do you know what derivatives are and how to use them? I suspect that you're are in Precalculus, and so you probably haven't learned Calculus yet. If this is true, say so and I, or one of the other helpers on this site, will show you how to do these problems without derivatives.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by Jhevon View Post
    hold on, you're in pre-calculus?

    that means you don't know how to find a derivative right?
    lol
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by ecMathGeek View Post
    juiicycouture, do you know what derivatives are and how to use them? I suspect that you're are in Precalculus, and so you probably haven't learned Calculus yet. If this is true, say so and I, or one of the other helpers on this site, will show you how to do these problems without derivatives.
    yeah, sorry, i realized afterwards that this thread was in precalc.

    we will need to use the vertex equation then
    Follow Math Help Forum on Facebook and Google+

  7. #7
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    ah well, juiicycouture went offline. you wanna do the precalc way anyway ecMathGeek? or should i? 'cause you seem bored
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    question 1
    let's start from

    so A = (8 - y)y = 8y - y^2

    for max, we want y = -b/2a .........this is the formula for the vertex of the parabola
    => y = -8/2(-1) = -8/-2
    => y = 4

    But x = 8 - y
    => x = 4

    question 2
    let's begin from here
    => A = (40 - 2y)y = 40y - 2y^2

    for max, y = -b/2a
    => y = -40/2(-2) = -40/-4
    => y = 10

    but x = 40 - 2y
    => x = 20
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by juiicycouture View Post
    1) a rectangular backyard playpen for child is to be enclosed with 16 m of fliexible fencing. what dimensions of the rectangle will provide the maximum area for the child to play?

    2) a rectangular corral is to be enclosed along the side of a horse barn with the barn serving as one side of the corral. what dimensions of the corral using 40 m of fencing, will enclose the maximum area for the horses?

    3) a rectangular garden plot requires an area of 32 m squared ...for the variety of vegetables that are tobe planted. what dimensions of the plot will use the lead amount of fencing to enclose the garden?
    3)
    The deminsions we will use are x = length, and y = width
    We are given A = 32 = xy
    We need P = 2x + 2y to be a minimum.

    First, let's solve 32 = xy for x (we could solve it for y, too... we'd get the same answer)
    32 = xy
    x = 32/y

    Second, plug in x = 32/y into P = 2x + 2y
    P = 2(32/y) + 2y
    P = 64/y + 2y
    P = 64y^(-1) + 2y ... since 1/y = y^(-1)

    Now, we need to make this a quadratic function (which we can do if we multiply everything by y)
    yP = 64 + 2y^2
    2y^2 - Py + 64 = 0

    For the minimum area, all we need is y = -b/(2a)
    y = -(-P)/(2*2)
    y = P/4

    Since P = 2y + 2x
    2x = P - 2y
    x = P/2 - y ... Plug in y = P/4
    x = P/2 - P/4
    x = P/4

    Therefore, x and y are one forth the total perimater.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by ecMathGeek View Post
    3)
    The deminsions we will use are x = length, and y = width
    We are given A = 32 = xy
    We need P = 2x + 2y to be a minimum.

    First, let's solve 32 = xy for x (we could solve it for y, too... we'd get the same answer)
    32 = xy
    x = 32/y

    Second, plug in x = 32/y into P = 2x + 2y
    P = 2(32/y) + 2y
    P = 64/y + 2y
    P = 64y^(-1) + 2y ... since 1/y = y^(-1)

    Now, we need to make this a quadratic function (which we can do if we multiply everything by y)
    yP = 64 + 2y^2
    2y^2 - Py + 64 = 0

    For the minimum area, all we need is y = -b/(2a)
    y = -(-P)/(2*2)
    y = P/4

    Since P = 2y + 2x
    2x = P - 2y
    x = P/2 - y ... Plug in y = P/4
    x = P/2 - P/4
    x = P/4

    Therefore, x and y are one forth the total perimater.
    I forgot to finish up this problem in solving for x and y (thanks Jhevon)

    Since y = P/4 and x = P/4, we can say y = x. We can also plug this relationship back into 32 = xy
    x^2 = 32
    x = Sqrt(32)
    AND
    y = Sqrt(32)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. optimization problems
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 25th 2009, 05:19 PM
  2. 2 Optimization Problems
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 5th 2007, 10:02 AM
  3. Optimization Problems
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 4th 2007, 08:20 PM
  4. Optimization Problems, 2
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 4th 2007, 10:25 AM
  5. Optimization problems
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 13th 2006, 11:55 PM

Search Tags


/mathhelpforum @mathhelpforum