Volume of a Box

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saaz

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Hello everyone,

I was doing this problem: the width of a rectangular box is 3 times its height, and its length is 11 inches more than its height. Find the dimensions of the box if its volume is 720 cubic inches.

Using the formula for a volume of a cube leads me to this (where h is the height of the box):
[math]h^3+11h^2-240=0[/math]
The problem here is that I don't know how to find h. I could use a graphic calculator, but is not specified in the problem and also I would like to be able to solve it analytically. Then I thought of using the rational root theorem to find all the possible roots but one power is missing (so I don't think I can use it simply by implying that the coefficient of h to the power of 1 is 0). I could use brute force and crunch some numbers till I get to 720, but I don't like so much that approach (if there is something better).

Has anyone a tip?

Thanks as always!

Saaz
 
Yes, try using the rational root theorem for h^3 + 11h^2 + 0h -240.
 
I meant using the rational root theorem implying that the coefficient (a of ah^1) is equal to zero, as you wrote: h^3 + 11h^2 + 0h -240. So is it possible to do this and use the rational root theorem? My book states that both the leading coefficient and the constant term must not be zero, but it doesn't specify the other coefficients and there is neither an example where it does the above.

It's still messy but I guess that's life...

Thanks!

Before I proceed with all the factors of 240 (there are a lot), what if I divide the volume by 8 as if all the three sides of the box are divided by 2? This way I should be able to find the value for h/2, but have to factor 30 instead of 240. I don't know, it just came to mind. I guess I'll try and see if it works.

Ah my bad, I can't. The leading coefficient must be an integer...
 
The reason I put in 0h was in case you are going to use synthetic division. Even if you use long division it would make it easier if you inserted 0h.

Why divide by just 8? You can divide by 240 (you already divided by 3). Then your constant would be 1 which only has two factors. The reason this can't be done is because the leading coefficient would be 1/30 (or 1/240). Now what are the factors of 1/30?? To use the rational root theorem the coefficients must be integers and that is why you can't divide by 8.
 
Assuming you were able to divide by 8, it is NOT true that your solution would tell you what h/2 equal. No! Again, you divided by 3 so when you find the solution would it be h/cube root (3) ? No, you will find h.

Example 2x-4 = 10. x=7 since 2*7-4=14-4=10
Now 2x-4 = 2(x-2) = 10. Dividing by 2 yields x-2=5. Then x=7, NOT x/2 = 7.
 
Another point. You have an equation h^3 + 11h^2 + 0h -240 and you want to find the root(s). Say that 7 is a root.
If all you are given is h^3 + 11h^2 + 0h -240, then I guess that you would say that h=7.
But what if h^3 + 11h^2 + 0h -240 came from a square where one side is called h. Is h=7 no longer a root?
But what if h^3 + 11h^2 + 0h -240 came from a rectangular box where one side is called h. Is h=7 no longer a root?
 
Last edited:
Yes, you are right... I was trying to think about the problem from a "physical perspective" but then I realised that I actually did the same by dividing by 3 (it's just that in my head that was arithmetic).
 
Steven G said:
Another point. You have an equation h^3 + 11h^2 + 0h -240 and you want to find the root(s). Say that 7 is a root.
If all you are given is h^3 + 11h^2 + 0h -240, then I guess that you would say that h=7.
But what if h^3 + 11h^2 + 0h -240 came from a square where one side is called h. Is h=7 no longer a root?
But what if h^3 + 11h^2 + 0h -240 came from a rectangular box where one side is called h. Is h=7 no longer a root?
Click to expand...
You are posting like our friend from *_nyc → 3 posts in 18 minutes........
 
saaz said:
Hello everyone,

I was doing this problem: the width of a rectangular box is 3 times its height, and its length is 11 inches more than its height. Find the dimensions of the box if its volume is 720 cubic inches.

Using the formula for a volume of a cube leads me to this (where h is the height of the box):
[math]h^3+11h^2-240=0[/math]
The problem here is that I don't know how to find h. I could use a graphic calculator, but is not specified in the problem and also I would like to be able to solve it analytically. Then I thought of using the rational root theorem to find all the possible roots but one power is missing (so I don't think I can use it simply by implying that the coefficient of h to the power of 1 is 0). I could use brute force and crunch some numbers till I get to 720, but I don't like so much that approach (if there is something better).

Has anyone a tip?

Thanks as always!

Saaz
Click to expand...
Let us assume that the value of 'h' is an integer - since all the given numbers are 'integers'. I CANNOT defend this assumption - but that is the most logical "simplifying" stepping stone towards solution.

With that in mind - we can see that h has to be less than 6 (←6^3 = 236) . So we need to test only 1,2,3,4,5.

With that knowledge we find that

h = 4 in
 
saaz said:
I was doing this problem: the width of a rectangular box is 3 times its height, and its length is 11 inches more than its height. Find the dimensions of the box if its volume is 720 cubic inches.

Using the formula for a volume of a cube leads me to this (where h is the height of the box):
[math]h^3+11h^2-240=0[/math]
The problem here is that I don't know how to find h. I could use a graphic calculator, but is not specified in the problem and also I would like to be able to solve it analytically. Then I thought of using the rational root theorem to find all the possible roots but one power is missing (so I don't think I can use it simply by implying that the coefficient of h to the power of 1 is 0). I could use brute force and crunch some numbers till I get to 720, but I don't like so much that approach (if there is something better).
Click to expand...
The rational root theorem only requires the coefficients to be integers, so there is no reason it wouldn't apply here. If you're not sure how to interpret it, show us how your textbook states it and we can point out the important features.

When I apply it, I generally just start with small divisors, both positive and negative, and work up; in this case the solution has to be positive. Testing 1, 2, 3, 4, 5, ..., you find the solution fairly quickly.

If you wanted assurance that you would definitely find the solution quickly, you could consider that h^3+11h^2 has to be 720, so h^3 must be less than 720. So the solution must be less than 10 (much less, since 10^3+11(10^2) = 2100), and you won't need to go through nearly the whole list. But that's generally assured sufficiently by the fact that a teacher assigned the problem to you ...

Generally, though there are formulas for solving cubics, this is far easier than using them, and is almost certain to work for school problems. In real life ... you'd use technology!
 
Thank you all for the comments! So, in the end, it's about reaching that first equation that I wrote and either using the rational zero theorem or crunching some numbers to get the result.
 
saaz said:
Thank you all for the comments! So, in the end, it's about reaching that first equation that I wrote and either using the rational zero theorem or crunching some numbers to get the result.
Click to expand...
Use the rational root test first!
 
This is overkill for this problem but I want to try Lagrange Multiplier (something I just learned). @saaz please ignore what I'm doing if you don't understand.

We seek to optimize [imath]f(h)=3h^3+33h^2[/imath], subject to the constraint [imath]3h^3+33h^2=720.[/imath]

The Lagrangian [imath]L(h,\lambda)=(3x^3+33x^2)+λ(3x^3+33x^2-720)[/imath]

[math] \begin{cases} \dfrac{\partial L}{\partial x}=3x(3x+22)(\lambda+1)=0\\ \\ \dfrac{\partial L}{\partial \lambda}=3x^3+33x^2-720=0\\ \end{cases} \implies \boxed{h=4,\lambda=-1} [/math]
Therefore the dimensions are [imath]4 \times 12 \times 15.[/imath]
 
AvgStudent said:
This is overkill for this problem but I want to try Lagrange Multiplier (something I just learned). @saaz please ignore what I'm doing if you don't understand.

We seek to optimize [imath]f(h)=3h^3+33h^2[/imath], subject to the constraint [imath]3h^3+33h^2=720.[/imath]

The Lagrangian [imath]L(h,\lambda)=(3x^3+33x^2)+λ(3x^3+33x^2-720)[/imath]

[math] \begin{cases} \dfrac{\partial L}{\partial x}=3x(3x+22)(\lambda+1)=0\\ \\ \dfrac{\partial L}{\partial \lambda}=3x^3+33x^2-720=0\\ \end{cases} \implies \boxed{h=4,\lambda=-1} [/math]
Therefore the dimensions are [imath]4 \times 12 \times 15.[/imath]
Click to expand...
Where did all those x's come from?
 
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