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Determining dimensions of a rectangle

  1. Oct 17, 2013 #1
    1. The problem statement, all variables and given/known data

    A box has a length that is 13cm longer than its width, and the volume of the box is 60cm^3. Determine the dimensions of the box.


    2. Relevant equations

    V = lwh
    l = 13cm > w
    h = ?

    3. The attempt at a solution

    Since V = lwh,
    60cm^3 = lwh and,
    l = 13cm > w
    w and h are unknown, but I'm not sure how to determine the other dimensions.
     
  2. jcsd
  3. Oct 17, 2013 #2
    It seems that the question is incomplete.Are you sure you haven't missed something ?
     
  4. Oct 17, 2013 #3
    That's all there is. The only other thing really is that it states the box has square ends.
     
  5. Oct 17, 2013 #4
    Well...That is a very important detail .Without this information you can not have a unique solution.

    Does this information tell you some relation between the width and the height of the square ?

    What property does a square have in terms of the length of its sides ?
     
  6. Oct 17, 2013 #5
    It may be only me, but what the hell did you wrote there?

    if length is 13cm longer than the width we could ADD 13cm to the width to get the length, id est:

    l = 13cm + w

    Maybe I look pesky now, but > is absolutely (at least according to my math studies) different from +.

    Tanya Sharma said a lot. Imagine how many different edge lengths you will have if the ends of the box are squares.
     
  7. Oct 18, 2013 #6
    They are different indeed. However, the length is 13cm GREATER THAN the width. It makes sense both ways, but saying the length is 13cm > the width is much easier to understand (for me anyways). Secondly, in terms of sides of a square, the property is that they are equal side lengths. But where would I go from there?
     
  8. Oct 18, 2013 #7
    My point is that you are going to want to set some equations and solve them, and you can't (logically) equal l to a comparison. (Tell me if this makes sense for you)

    A box (parallelepiped) has 3 different edge lengths (or less), so...

    You already know that l1 = l2 + 13

    Now try to get a l3 from that square property you told me.

    As we have ONE equation, we can solve to ONE variable.

    We know that l1l2l3 = 60
    We need to have the aforementioned equation in terms of l1 or l2 or l3
     
  9. Oct 18, 2013 #8
    You need to express the relationship between length and width mathematically i.e via an equation .So what is it ?
     
  10. Oct 18, 2013 #9
    Okay, I understand what you meant by that now; it makes sense that one cannot have l equal to a comparison. I'm unsure of determining l3 from the square property though.
     
  11. Oct 18, 2013 #10
    A = lw
    and
    V = lwh
     
  12. Oct 18, 2013 #11
    Please reply to post#8
     
  13. Oct 18, 2013 #12
    I did.
    A = lw and V = lwh
     
  14. Oct 18, 2013 #13
    This doesn't make any sense to what I asked.

    The length is greater than the width by 13 cm .How do you express this relation mathematically via an equation ?
     
  15. Oct 18, 2013 #14
    Wow. How about you specify more clearly like you just did first, that way I understand exactly what you are asking the first time. Anyway,

    l = 13cm + w
    Correct?
     
  16. Oct 18, 2013 #15
    Good...Now look at the attachment.One of the square ends is painted yellow .Does this tell you something about the width and height of the box?
     

    Attached Files:

  17. Oct 18, 2013 #16

    Mark44

    Staff: Mentor

    Mostly. You can omit the units (cm). All dimensions here are in the same units, so it's not necessary to carry them along. Also, it makes no sense to include units for 13 but not for w.
     
  18. Oct 19, 2013 #17
    Okay, yes! I figured it out last night. The width and the height will be equal. Thank you, Tanya Sharma, besulzbach, and Mark44!

    Therefore, since V = 60cm^3 then w = 2 cm and h = 2 cm and thus l = 15cm
     
  19. Oct 19, 2013 #18
    Exactly.

    Notice that if you try to "proof" it mathematically you could do the following:

    [itex]l_{1} \cdot l_{2} \cdot l_{3} = 60[/itex]

    [itex]l_{1} \cdot (l_{1} + 13) \cdot l_{3} = 60[/itex]

    [itex]l_{1} \cdot (l_{1} + 13) \cdot l_{1} = 60[/itex]

    [itex]l_{1}^{2} \cdot (l_{1} + 13) = 60[/itex]

    [itex]l_{1}^{3} + 13l_{1}^2 = 60[/itex]

    [tex]l_{1}^{3} + 13l_{1}^2 - 60 = 0[/tex]

    Now you would solve that cubic equation and get its roots. (You could try it as an exercise)
     
  20. Oct 19, 2013 #19

    Mark44

    Staff: Mentor

    The word is "prove".

    The equations below are more complicated than they need to be. It's given in the problem that two of the dimensions are equal, so you don't need three variables.

    Although it is possible to solve cubic equations, doing so can be very complicated. A better approach is what was done earlier in this thread.
     
  21. Oct 19, 2013 #20
    Yes, of course. Thank you. I'm not a native English speaker, I messed it up.

    What exactly was done?

    We got l, w and h in terms of one variable. Just getting {2, 2, 15} through "thinking" may not satisfy some tutors/mentors/teacher/evil cyborgs.

    I don't know if it would be possible to solve this problem with a quadratic equation.
     
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