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Determining dimensions of a rectangle

  • Thread starter neuro.akn
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  • #1
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Homework Statement



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.


Homework Equations



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

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.
 

Answers and Replies

  • #2
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Homework Statement



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.


Homework Equations



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

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.
It seems that the question is incomplete.Are you sure you haven't missed something ?
 
  • #3
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That's all there is. The only other thing really is that it states the box has square ends.
 
  • #4
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134
That's all there is. The only other thing really is that it states the box has square ends.
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 ?
 
  • #5
28
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A box has a length that is 13cm longer than its width


l = 13cm > w
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.
 
  • #6
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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?
 
  • #7
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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
 
  • #8
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You need to express the relationship between length and width mathematically i.e via an equation .So what is it ?
 
  • #9
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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.
 
  • #10
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A = lw
and
V = lwh
 
  • #11
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Please reply to post#8
 
  • #12
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I did.
A = lw and V = lwh
 
  • #13
1,540
134
I did.
A = lw and V = lwh
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 ?
 
  • #14
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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?
 
  • #15
1,540
134
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?
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?
 

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  • #16
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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?
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.
 
  • #17
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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
 
  • #18
28
1
Therefore, since V = 60cm^3 then w = 2 cm and h = 2 cm and thus l = 15cm
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)
 
  • #19
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Exactly.

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

[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)
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.
 
  • #20
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The word is "prove".
Yes, of course. Thank you. I'm not a native English speaker, I messed it up.

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.
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.
 
  • #21
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5,192
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.
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.
You're right. The OP didn't say how he/she came up with the solution. In this case, solving the cubic is the way to go.
 
  • #22
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It's given in the problem that two of the dimensions are equal, so you don't need three variables.
I wan't further explanation on this.

Well, we can say that [itex]l_{1} = l_{2}[/itex] and then use [itex]l_{1}^{2}[/itex] to account for both variables.

But, when you say that a third variable is not necessary you are claiming what exactly? Do you see a way to get a quadratic equation for this problem?

I don't mean to be rude, I just want to know if there is a simpler way because some problems I get at school are really similar and solving quadratics is, as you said before, usually considerably faster (and easier).
 
  • #23
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I wan't further explanation on this.
The OP didn't think this was important enough to put in the original post:
That's all there is. The only other thing really is that it states the box has square ends.
Well, we can say that [itex]l_{1} = l_{2}[/itex] and then use [itex]l_{1}^{2}[/itex] to account for both variables.
Instead of using one variable with subscripts, why not use different variables whose names are reflective of what they represent? IOW, l for length, w for width, and h for height.

Based on what I think is the level of mathematical sophistication, names like l, w, and h are easier to understand than l1, l2, and l3. Furthermore, the notation gets pretty gnarly when you combine subscripts and exponents as you did; e.g. l12. Here the notation becomes an impediment to understanding.
But, when you say that a third variable is not necessary you are claiming what exactly? Do you see a way to get a quadratic equation for this problem?
The given information in the problem is that the end is square, so h = w, and that the length is 13 more than the width, so l = 13 + w.

From this we can write the equation for volume as 60 = l * w * h = (13 + w) * w2. This is a cubic, of course. To find the dimensions you have to solve this cubic equation.
I don't mean to be rude, I just want to know if there is a simpler way because some problems I get at school are really similar and solving quadratics is, as you said before, usually considerably faster (and easier).
Right, and I don't consider you asking to be rude. The equation you end up having to solve is a cubic - there's no getting around that. Sometimes cubics are difficult to solve, but in this case, using the rational root theorem and synthetic division (or long division if you don't know synthetic division), one root pops out almost immediately.
 
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