Solve Area Minimization Question: Tent w/o Bottom & Triangles

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dobedobedo
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I've got a question that I don't know how to solve. The question is:

We want to produce a tent, without a bottom part, which has two rectangular sides and two gables in the form of two isosceles triangles with the base against the ground. Determine the height of the tent, which has volyme V and requires the least amount of cloth.



And by the "amount of cloth" I believe that they mean the area which the surfaces of the tent occupy. The answer is [itex](\frac{V}{\sqrt{2}})^{1/3} = h[/itex], but I don't know how to get to this answer. I guess that one should first try to find the expression of the surface area, and then try to optimize it under some equality constraint. My guess is that the iscoseles triangle is some sort of equality constraint... but I don't know much more.

I can haz this question explained pweez?
 
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hi dobedobedo! :wink:

start by giving things names

call the length of the triangle "a" and the length of the rectangular side "b"

then write out the formula for the area of the cloth …

what do you get? :smile:​
 
Been there done that. I don't get anything.
 
Give it a shot man/dude/girl! I'm in pain, i really want to fix this problem.
 
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Let a be the side of that isosceles triangle and b be the length of the rectangle and let h be the height of that triangle. Then the area A is:

[itex]A(a,b,h) = 2ab + 2*h*\sqrt{a^2-h^2}[/itex].
Great. Names. Woohoo. Now what?
[itex]V = h*b*\sqrt{a^2-h^2}[/itex].
 
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I've been working on this too - no solution.
Calculated the area (use 2b for the base, h is height, l length)
A = 2(hb + l√(h2 + b2))
and differentiated wrt h to find a minimum, I get a result that I can't bring down to the given answer.
b= -lh/√(h2 + b2)
 
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dobedobedo said:
Let a be the side of that isosceles triangle and b be the length of the rectangle and let h be the height of that triangle. Then the area A is:

[itex]A(a,b,h) = 2ab + 2*h*\sqrt{a^2-h^2}[/itex].
Great. Names. Woohoo. Now what?
[itex]V = h*b*\sqrt{a^2-h^2}[/itex].
I suggest eliminating b, and writing a2 = h2+w2
[itex]b = \frac{V}{h*w}[/itex]
[itex]A = \frac{2\sqrt{h^2+w^2}V}{h*w} + 2*h*w[/itex]
Note that this is symmetric in w and h, so minimising wrt one will give the same answer as wrt the other. So we can assume h = w. That makes the differentiation a little easier and appears to produce the desired answer.
 
dobedobedo said:
Let a be the side of that isosceles triangle and b be the length of the rectangle and let h be the height of that triangle. Then the area A is:

[itex]A(a,b,h) = 2ab + 2*h*\sqrt{a^2-h^2}[/itex].
Great. Names. Woohoo. Now what?
[itex]V = h*b*\sqrt{a^2-h^2}[/itex].

You really need to show more effort on your schoolwork questions here. Trying to be cute while showing so little effort will not get you far...

Check your PMs please.
 
haruspex said:
I suggest eliminating b, and writing a2 = h2+w2
[itex]b = \frac{V}{h*w}[/itex]
[itex]A = \frac{2\sqrt{h^2+w^2}V}{h*w} + 2*h*w[/itex]
Note that this is symmetric in w and h, so minimising wrt one will give the same answer as wrt the other. So we can assume h = w. That makes the differentiation a little easier and appears to produce the desired answer.

Hm. Thanks for the help, but that's strange. It does produce the desired answer. In this case we've got two variables, so I think it is okay to make such an assumption. If it were the case of more variables, would such an assumption be okay if the expression were symmetric w.r.t. two particular variables?
 
dobedobedo said:
Hm. Thanks for the help, but that's strange. It does produce the desired answer. In this case we've got two variables, so I think it is okay to make such an assumption. If it were the case of more variables, would such an assumption be okay if the expression were symmetric w.r.t. two particular variables?
If the expression is entirely symmetric wrt two variables, whatever you can deduce regarding one must also be deducible re the other. That said, complications could arise. E.g. suppose the vars are x, y and you deduce (not assuming x=y) that x=±1. The same will be true for y, but maybe x and y have to have opposite sign. So plugging in x=y won't give the right result. Having written out that response, I now regard my trick as rather untrustworthy :blushing:.