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Thread: Need math help. =] +rep.

  1. #1
    Queen of Candy Land Ray-Chill's Avatar
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    Need math help. =] +rep.

    Could anyone explain how to do this?

    A box with a square base and no top is to be made from a square piece of carboard by cutting 5 in. squares from each corner and folding up the sides. The box is to hold 9245 in3. How big a piece of cardboard is needed?

    I think I've got the basics.. but I'm getting wrong answers..

    and also,
    You have a wire that is 68 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a square. The other piece will be bent into the shape of a circle. Let A represent the total area of the square and the circle. What is the circumference of the circle when A is a minimum?
    Last edited by Ray-Chill; 02-29-2012 at 12:29 PM. Reason: Added another hard ass problem.

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    SmileYaDead's Avatar
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    I'll give it a try, seems rather easy..

    I think the first one is like this:

    V=a*b*c

    We know a=b

    We know c=5

    We know V=9245

    9245=a*b*5

    5ab=9245 / 5

    ab=1849 (sqrt)

    a=b=43

    Since you cut 5 inches from each side, to form the edges for the box, the original piece has to be 10 inches wider .

    43+10=53

    P=53*53=2809 squareinches.
    Last edited by SmileYaDead; 02-29-2012 at 03:15 PM.

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    Ray-Chill (02-29-2012)

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    edit: already answered >:
    Last edited by Tom; 02-29-2012 at 03:41 PM.

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    SmileYaDead's Avatar
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    The second one is a real b*tch, but I'll try and think of something.

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    Slasher's Avatar
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    For the second number
    64/4 = 16
    16 for each side of the square
    4 left to make the circle

    I don't really understand what you need to do...

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    SmileYaDead's Avatar
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    Slasher, does it guarantee the smallest total area?

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    bamag's Avatar
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    I think I did the exact same problem for my precalc class last term. I'll give it a shot

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    Slasher is right about the numbers, we just need a decent way to show it.

    Square:

    P=a*b, a=b

    Circle:

    A=PI*r^2

    r=d/2 (radius, diameter)

    C/d=PI; C=d*PI; d=C/PI (circumference)

    Let's say the square's side equals 16. 4*16=64; 68-64=4 ;

    So, C=4

    Since d=C/PI, d=4/PI

    r = d/2 = 4/PI/2 = 2/PI

    A = PI*r^2 = PI*(2/PI)^2 = PI*(4/PI^2) = 4PI/PI^2 = 4/PI is the circle's area. Square's area=16*16=256.
    Total area=256+(4/PI)=256+1.2732395447351626861510701069801=257.2732 395447351626861510701069801



    To show that this is the smallest possible, we must take as big circumference of the circle as we can. So, we take 64, since we need 1 for each side of the square.

    P=1*1=1

    C=64

    d=64/PI

    r=64/PI/2=32/PI

    A=PI*(32/PI)^2 = PI*(1024/PI^2) = 1024PI/PI^2=1024/PI

    The total are would now be 1+1024/PI, which is about 326.94932345220164765467394738691.

    Since 257. *** is smaller than 326. *** we can now see the smallest possible value of A is gotten with the smallest possible value of the circumference, which is 4.

    Works?
    Last edited by SmileYaDead; 02-29-2012 at 05:21 PM.

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    bamag's Avatar
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    I think I did the exact same problem for my precalc class last term. I'll give it a shot
    You have a wire that is 68 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a square. The other piece will be bent into the shape of a circle. Let A represent the total area of the square and the circle. What is the circumference of the circle when A is a minimum?
    Let l be 1 side of square.
    Total length of square = 4l.
    Circumference of circle = 68-4l.
    68-4l=2r*pi
    r=(68-4l)/(2pi)=(34-2l)/pi
    Area of square = l^2
    Area of circle = pi((34-2l)/pi)^2 = pi(4l^2-136l+1156)/(pi^2)=(4l^2-136l+1156)/pi
    A=l^2+(4l^2-136l+1156)/pi=(pi*l^2+4l^2-136l+1156)/pi
    Now you have two choices.
    Now you have a quadratic and there are two ways of solving for the minimum length.
    If you are taking calculus, you take the derivative and set it equal to 0.
    As I have not taken calculus yet (taking precalc at the moment), I'll solve it without calculus.
    The minimum value of A would be at the axis of symmetry.
    The axis of symmetry is l=-b/2a with b=-136/pi and a=(4+pi)/pi.
    l=136/(2(4+pi))=68/(4+pi)
    Now you solve for circumference:
    68-4l=68-4(68/4+pi)=68-272/(4+pi)=(272+68pi-272)/(4+pi)=68pi/(4+pi)
    Therefore the circumference when A is minimum is 6pi/(4+pi) cm which is around 2.64cm.
    This should be correct assuming I didn't mess up somewhere...
    Whew that was a lot of work lol :p
    Can anyone confirm/refute my solution?

    Oh and calculus solution:
    Derivative of A=(pi*l^2+4l^2-136l+1156)/pi is (2(-68+4l+pi*l))/pi
    You set that to 0:
    (2(-68+4l+pi*l))/pi=0
    -136+4l+pi*l=0
    4l+pi*l=136
    2l=136/(4+pi)
    l=68/(4+pi)
    Rest is the same as other solution.

    @smileyadead The total length of the square does not have to be 64 so I don't really see why that would be the solution.
    Last edited by bamag; 02-29-2012 at 06:07 PM.

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  13. #10
    SmileYaDead's Avatar
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    I figured it from the first one, which was rather easy and dealt with full numbers. Why would they give one that easy and the other one hard as heck? I don't really know what grade she's in tho

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