Results 1 to 6 of 6

Math Help - I think this is calc anywyas..

  1. #1
    Newbie
    Joined
    Dec 2006
    Posts
    10

    I think this is calc anywyas..

    Im on some homework and came across this problem.

    A contractor is to fence off a rectangular field along a straight river, the side along the river requiring no fence

    a) What is the least amount of fencing needed to fence off 30,000m^2
    I'm assuming im supposed to use 2X + Y = 30,000m^ for perimiter. However i know that i cant have two variables in the function. What'd be the function from here?

    and b) What is the greatest area the contractor can fence off using 500m of fencing.
    This is trying to find the maxima for area... and the formula would be (x^2)(y) for Volume yes? how do you get the function from here? Im pretty sure I can finish it after i get the function though

    __________________________________________________ ___--

    And theres another problem..
    dont remember how its worded but it wants the minimum of the sum of two positive numbers, where the first is ^3 and the second is ^2. Also, they, without the ^, = 4
    so
    X+Y=4
    Y= (-x+4)
    X^3 + (-x+4)^2 -> X^3 + X^2 -8X +16
    F'= 3x^2 + 2x -8

    and after using -b plus/minus (Radical b^2 (-4)(a)c) divided by 2a
    the x's are .. 1.33333333333etc and .. -2 i think

    and the 1.3333333333333 comes to be the min because after plugging it into f" = 6x +2 it is a Positive concave,

    so 1.3333333333 being x, i'd put it into f' to find the y, which comes to be.. around 9.45 or something similar,
    and putting that into the origional equation, (1.33333333^3) + (9.45^2) = around 95 or a high number like that


    HOWEVER this MUST be wrong because 9.45 + 1.3333 does NOT = 4

    Where did i mess up on this one?




    thanks for your help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    Re:

    RE:

    Go the following link => http://www.mathhelpforum.com/math-he...20-2007-a.html and see #4. Look around the forum as we have already solved millions of these optimization problems before.

    -qbkr21
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,923
    Thanks
    332
    Awards
    1
    Quote Originally Posted by aargh27 View Post
    A contractor is to fence off a rectangular field along a straight river, the side along the river requiring no fence

    a) What is the least amount of fencing needed to fence off 30,000m^2
    I'm assuming im supposed to use 2X + Y = 30,000m^ for perimiter. However i know that i cant have two variables in the function. What'd be the function from here?
    No. 2X + Y is the perimeter, or length, of the fence.
    X*Y = 30000.
    P = 2X + Y

    You wish to minimize P. Thus:
    Y = 30000/X ==> P = 2X + 30000/X

    dP/dX = 2 - 30000/X^2 = 0

    2 = 30000/X^2

    X^2 = 30000/2 = 15000

    X = 122.47 m

    Is this a minimum point? Well,
    d^2P/dX^2 = -60000/X^3 > 0 for X = 122.47
    Thus this is indeed a minimum point. I'll let you decide if this is merely a relative minimum or an absolute minimum.

    X = 122.47 m ==> Y = 30000/122.47 = 244.95 m

    So
    P = 2[122.47] + [244.95] = 489.90 m

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,923
    Thanks
    332
    Awards
    1
    Quote Originally Posted by aargh27 View Post
    A contractor is to fence off a rectangular field along a straight river, the side along the river requiring no fence

    b) What is the greatest area the contractor can fence off using 500m of fencing.
    This is trying to find the maxima for area... and the formula would be (x^2)(y) for Volume yes? how do you get the function from here? Im pretty sure I can finish it after i get the function though
    Volume?? You're looking for area!

    Try this on your own. It's basically a redo of part a) except now you know the perimeter and you are trying to maximize the area.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,923
    Thanks
    332
    Awards
    1
    Quote Originally Posted by aargh27 View Post
    And theres another problem..
    dont remember how its worded but it wants the minimum of the sum of two positive numbers, where the first is ^3 and the second is ^2. Also, they, without the ^, = 4
    so
    X+Y=4
    Y= (-x+4)
    X^3 + (-x+4)^2 -> X^3 + X^2 -8X +16
    F'= 3x^2 + 2x -8

    and after using -b plus/minus (Radical b^2 (-4)(a)c) divided by 2a
    the x's are .. 1.33333333333etc and .. -2 i think

    and the 1.3333333333333 comes to be the min because after plugging it into f" = 6x +2 it is a Positive concave,

    so 1.3333333333 being x, i'd put it into f' to find the y, which comes to be.. around 9.45 or something similar,
    and putting that into the origional equation, (1.33333333^3) + (9.45^2) = around 95 or a high number like that


    HOWEVER this MUST be wrong because 9.45 + 1.3333 does NOT = 4

    Where did i mess up on this one?
    What the heck are you doing in the section in red?

    And if they want the minimum sum of the two numbers how do you know it is 4?

    Okay, assuming your setup is correct you want the minimum of the function
    F = X^3 + X^2 - 8X +16

    So
    dF/dX = 3X^2 + 2X - 8 = 0

    X = -4/3, 2

    For which value of X is F a minimum?

    Well,
    d^2F/dX^2 = 6X + 2

    For X = -4/3, d^2F/dX^2 = 6*(-4/3) + 2 = -6 < 0 so this represents a local maximum.

    For X = 2, d^2F/dX^2 = 6*2 + 2 = 14 > 0 so this represents a local minimum. (Again I'll let you decide if this is an absolute minimum.)

    So X = 2 ==> Y = -2 + 4 = 2, so the two numbers are 2 and 2.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Dec 2006
    Posts
    10
    And if they want the minimum sum of the two numbers how do you know it is 4?
    The X + Y = 4 was given in the question.

    what i want the min for, would be X^3 + X^4

    for the part in red i didnt have the question with me, it was one i did earleir and was trying to remember it

    however i think what i did wrong, was i messed up the negative and positive signs partway instead of -4/3, 2, i had 4/3, -2. <.<


    Volume?? You're looking for area!
    gah ive been doing too much today =P
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Calc 1 review for Calc 2
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 28th 2011, 03:47 AM
  2. A few Pre-Calc problems in my AP Calc summer packet
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: August 30th 2010, 03:40 PM
  3. Replies: 1
    Last Post: January 13th 2010, 11:57 AM
  4. What am I doing? Calc, Pre-calc, Trig, etc...
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: April 23rd 2008, 08:51 AM
  5. Adv Calc lim prf
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 6th 2007, 09:25 AM

Search Tags


/mathhelpforum @mathhelpforum