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Optimization Problem/application of the derivative

  1. Jan 5, 2012 #1
    1. The problem statement, all variables and given/known data

    A wire 6 meters long is cut into twelve pieces, eight of one length and four of another. These pieces are welded together at right angles to form the frame of a box with a square base.

    a) Where should the cuts be made to maximize the volume of the box?
    b) Where should the cuts be made to maximize the are of the box?

    2. Relevant equations

    The derivative in relation to volume and area.
    Perimeter=6meters
    Perimeter=8x+4y=6meters
    x=(4y-6)/8
    y=(6-8x)/4
    A(x)=4xy + 2x^2
    V(x)=x^2y
    A'(x)=8x + 4y
    V'(x)=2xy + x^2

    where some 'x' and some 'y' are of some meters and less than 6 meters.

    3. The attempt at a solution

    So, I understand that setting the derivative of a function equal to zero of a function can allow someone to find the maximum and minimum of that function which is very practical in many areas, however, I can't seem to find the solution to this problem. It appears that the solution doesn't exist, but I must be doing something wrong.

    Here's what I did for a)

    a)
    since V'(x)=0 when volume is optimized, then
    0=2xy + x^2
    y=(6-8x)/4
    0=(2x(6-8x)/4) + x^2
    -x^2=(6x/2) - (4x^2)
    3x^2=(6x/2)
    x= 1 meter
    oh I just solved it,
    but if x=1meter, then y=((6-8(1))/4)meters=-1/2meters, but how is this quantity a negative? This is where I get stuck. I must be doing something wrong.

    b)
    since A'(x)=0 where area is optimized
    0=8x + 4y
    y=(6-8x)/4
    8x=-4(6-8x)/4
    32x=-24+32x
    x=0? o.0??? can someone show me what I'm doing wrong?
     
  2. jcsd
  3. Jan 5, 2012 #2

    ehild

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    Gold Member

    What did you mean on A' and V'? Anyway, your procedure is not correct. You can not ignore the derivative of y with respect to x.

    Substitute y=(6-8x)/4 for y both in A and V, and find the derivative with respect to x.

    ehild
     
  4. Jan 5, 2012 #3
    Oh ok! BTW, you DO know what I mean by A' or V' since it represents the derivative, but I realize that it is not consistent with the chain rules involved by saying dy/dx, and so forth, but that's another thing I learned and thanks! here's the correction then:

    V(x)=x^2(6-8x)/4
    V(x)=(6x^2/4) - (8x^3/4)
    V'(x)=(3x) - (6x^2)
    0=(3x) - (6x^2)
    x=.5meters

    and if 6meters=8x + 4y
    (6meters) - (4meters)=4y
    y=.5meters

    Volume=x^2y & V=(.5)^2(.5)m^3=.125m^3

    OHHHH Ok, that makes a lot of sense now. At first, somewhere around my calculations, I came up with this on my paper, but I didn't trust it because it looked so small, but it makes sense now. That truly is the maximum volume achieved. Ok, so on to the next question.

    b)

    since A'(x)=0 where area is optimized
    A(x)=4xy + 2x^2
    y=(6-8x)/4
    A(x)=4x(6-8x)/4 + 2x^2
    A(x)=(24x-32x^2)/4 +(2x^2)
    A(x)=6x - 8x^2 + 2x^2
    A(x)=6x - 6x^2
    A'(x)=6 - 12x
    0=6-12x
    x=.5 meters

    y=(6-8(.5))/4
    y=.5meters

    Oh! ok got them both! Thanks my friend. Much MUCH appreciated. So, another thing I want to elaborate on is the idea of the chain rule and how it applies to these kind of problems. So, in rates of change, I noticed that we use Leibniz notation a lot, because of the chain rule and we could do the same with optimization problems. The only reason why we don't do that is because one dimension usually relates in some form to another dimension like in this problem. I love that you pointed that out. Makes me think more about what I'm doing. :) I have another question coming up on how you use the derivative of a function of theta to find the rate of theta at any given dimensional point. PLEASE stay tuned! :)
     
  5. Jan 5, 2012 #4

    ehild

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    Homework Helper
    Gold Member

    You are welcome. I watch your threads :smile:

    ehild
     
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