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There is a solid with triangular prism of fixed volume and a regular tetrahedron at each end.The surface area is given by A(x)= 3sqr3/2(x^2+16/x), where x is the length of each side of the tetrahedron.
Find the value of x so that the solid has a minimum surface area.
Thanks in advance.

Comments

( 5 comments — Leave a comment )
jkrissw
Oct. 31st, 2006 03:49 pm (UTC)
x = a whole bunch
thisgirliknow
Oct. 31st, 2006 03:49 pm (UTC)
you are oh so very wise, not to mention extremely helpful.

jkrissw
Nov. 1st, 2006 05:38 pm (UTC)
Glad to be of assistance!
tevarin
Oct. 31st, 2006 04:32 pm (UTC)
LJ needs a Math font
That's 3[(3/2)^(0.5)](x^2 + 16/x), yes? If I've transcribed the formula wrong, let me know. Puzzling out the text string is a pain.

It's a cool problem. The quick answer is that the triangular prism should be moderately short and fat. As the prism gets very long and skinny, x gets small, and the 16/x term gets very big. But if the prism gets too short (and thus too fat) then x becomes large and the x^2 term gets too big.

More precisely, I'd say the way to do it is to take the derivative of A(x) with respect to x, and look for points where it goes to zero, then evaluate A(x) at each of those points. The smallest A(x) value should be the minimum surface area.

So we're looking for the minimum value of 3[sqr(3/2)](x^2 + 16/x). Taking the derivative, we get A'(x) = 3[sqr(3/2)](2x - 16/(x^2))

there's probably an exact way to find the roots (aka points where A'(x) goes to zero). But I can't think of it, so I'll trial-and-error it.

For very small positive X, 2x - 16/(x^2) = large negative numbers.
If x = 0.5, 2x - 16/(x^2) = -63
If x = 1, 2x - 16/(x^2) = -14
If x = 1.5, 2x - 16/(x^2) = -4.11
If x = 2, 2x - 16/(x^2) = 0, Yay!
If x = 2.5, 2x - 16/(x^2) = 2.44
If x = 3, 2x - 16/(x^2) = 4.22
For larger positive X, 2x - 16/(x^2) = large positive numbers.

So the only real, positive root for the derivative is x=2, which means the area is 3[(3/2)^(0.5)](2^2 + 16/2) = 3[(3/2)^(0.5)](12) = ~44.1 square units, for a tetrahedron side length of 2 units.

If we compare, for x = 2.1, area = 3[(3/2)^(0.5)](2.1^2 + 16/2.1)
= 3[(3/2)^(0.5)](12.029)

For x = 1.9, area = 3[(3/2)^(0.5)](1.9^2 + 16/1.9)
= 3[(3/2)^(0.5)](12.031)

So it indeed looks like x = 2.0 gives a local minimum in surface area, and we can be reasonably sure that this is the overall minimum as well.

tevarin
Oct. 31st, 2006 05:32 pm (UTC)
Re: LJ needs a Math font
Wait a second. If my 44.1 square unit answer above is correct, then that's an awfully long, skinny prism. 5 units or so. that can't be right. I think there might be a problem with the way I wrote the area formula.

Exposed surface area of the prism = 3XL, where X is the edge length of the tetrahedron, and L is the length of the prism. The volume of the prism must be a constant, which means X^2 * L must equal a constant C, so
L = C/(x^2), so exposed area = 3C/X

Exposed surface area of the two regular tetrahedra = 6 * triangle area = 6 * (3^0.5)/4 * x^2

So total exposed area is 3C/x + (6/4 * (3^0.5) * x^2).

Combining terms, that's 3/2 * 3^0.5 * (x^2 + (2C/((3^0.5)* x))

If we assume 2C/(3^0.5) = 16, then the total exposed surface area is
3/2 * 3^0.5 * (x^2 + 16/x).

So for x = 2, the surface area is 3/2 * 3^0.5 * 12 = 31.2 square units, which is friendlier, and gives a shorter, fatter prism 3.5 units long.
( 5 comments — Leave a comment )

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thisgirliknow
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