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Cuboid

cinelli
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Cuboid

#22117

Postby cinelli » January 11th, 2017, 12:26 pm

A cuboid is constructed so that

1) it has a volume of 7.5 cubic cms
2) its total surface area is 26 square cms
3) the total length of all its edges is 26 cms.

What are the cuboid’s dimensions?

Cinelli

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Re: Cuboid

#22209

Postby psychodom » January 11th, 2017, 4:06 pm

let dimension of cuboid be a, b, c

1. abc = 7.5
2. 2(ac + ab + bc) = 26
3. a + b + c = 6.5

If we double all dimensions:
4. ABC = 60 (ie x8)
5. AC + AB + BC = 52 (ie x4)
6. A + B + C = 13 (ie x2)

then I reckon ABC are going to be integers (sum and product of A,B,C is integer)

looking at factors of 60 and checking their sums gives
(2,3,4,5,6,10,12,15,30)

2+3+10 = 15
2+5+6 = 13 <--
3+4+5 = 12

therefore 2, 5, 6 work for A, B, C
so 1, 2.5, 3 work for a, b, c
and checking, this works for 2(ab + ac + bc) = 26, as required

-Dom

PS this argument seems very woolly, sorry!

Gengulphus
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Re: Cuboid

#22314

Postby Gengulphus » January 11th, 2017, 11:52 pm

cinelli wrote:A cuboid is constructed so that

1) it has a volume of 7.5 cubic cms
2) its total surface area is 26 square cms
3) the total length of all its edges is 26 cms.

What are the cuboid’s dimensions?

I'm still rather uncertain about the value of spoiler separators on these boards, but one won't do any real harm...

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Let the three dimensions in centimetres of the cuboid be a, b and c.

Then:

volume = abc = 7.5

area of all faces = 2(ab+bc+ca) = 26, so ab+bc+ca = 13

length of all edges = 4(a+b+c) = 26, so a+b+c = 6.5.

Now consider the cubic equation:

(x-a)(x-b)(x-c) = 0

Clearly its three roots are a, b and c. But (x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc = x^3 - 6.5x^2 + 13x - 7.5.

So the dimensions of the cuboid are the three roots of x^3 - 6.5x^2 + 13x - 7.5 = 0. So just apply the standard formulae for solving a cubic and there's the solution...

What? You don't know the standard formulae for solving a cubic??? Shame on you!

And shame on me as well - I don't know them either! I can look them up - but I never do, because they're messy and not very informative... When faced with a cubic equation, I either try to spot a solution - if I get one, I can factor it out and be left with a quadratic equation, which I do know the standard formula for - or I solve it numerically.

In this case, it's quite easy to spot that x=1 is a solution, since it causes x^3 - 6.5x^2 + 13x - 7.5 to be 1 - 6.5 + 13 - 7.5 = 0. Factoring it out, I get x^3 - 6.5x^2 + 13x - 7.5 = (x-1)(x^2-5.5x+7.5), so the other two dimensions are the roots of the quadratic equation x^2-5.5x+7.5 = 0, which are (5.5+/-sqrt(5.5^2-4*1*7.5))/2 = (5.5+/-sqrt(30.25-30))/2 = (5.5+/-0.5)/2 = 3 or 2.5.

So the dimensions of the cuboid are 1, 2.5 and 3.

Gengulphus

cinelli
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Re: Cuboid

#24677

Postby cinelli » January 20th, 2017, 11:37 am

Psychodom beat Gengulphus to the correct solution with his intuitive approach. If the unknowns a, b and c, he writes the conditions as

ABC = 60, BC + AC + AB = 52 and A + B + C = 13

where A = 2a, etc. He assumes that A. B and C must be integers. But is this necessarily the case? Consider the example where

LMN = 2, MN + LN + LM = 6, L + M + N = 5

In this case it turns out that the solution is

L, M, N = 1, 2+sqrt(2), 2-sqrt(2)

So here, it turns out that despite the equations involving all integers, the solution does not.
However since Psychodom found an integral solution to the problem and because we know that we are solving a cubic which has a unique solution, his solution must be right.

Cinelli

Gengulphus
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Re: Cuboid

#24900

Postby Gengulphus » January 21st, 2017, 8:43 am

cinelli wrote:Psychodom beat Gengulphus to the correct solution with his intuitive approach.

Well, actually my approach was intuitive as well. A different intuition to his: mine was that I might be able to spot one of the cubic's three solutions and actually spotting it, his was that values A, B and C that satisfy ABC = 60, BC + AC + AB = 52 and A + B + C = 13 might be integers.

What is true is that my solution contained more formal mathematics (so I suppose you could call his solution "more intuitive"). As a result, I got not just the solution, but also its essential uniqueness ("essential" because there are actually six solutions for (A,B,C), namely the three roots of the cubic in each of their six possible orders).

But of course, the crucial factor in him getting the solution first had little or nothing to do with intuitiveness - when people happen to read the board is a much more major factor!

Gengulphus


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