yashar806
I need help, please
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yashar806
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I have attached the question
anonymous
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we can do this
yashar806
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yes, could you explain it ?
anonymous
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ok first off z is 4, so lets forget z and replace it by 4
anonymous
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the volume of the first tube, the one with side \(x\) is \(4\times (\frac{1}{4}x)^2\)
yashar806
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why is it 1/4 ?
anonymous
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because your entire length is \(x\) and you are folding it in to a square
so the side of the square is \(\frac{1}{4}x\)
yashar806
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ok
anonymous
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and therefore the area of the square is \((\frac{1}{4}x)^2=\frac{x^2}{16}\) and since \(z=4\) the volume is \(4\times \frac{x^2}{16}=\frac{x^2}{4}\)
yashar806
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ok?
anonymous
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the volume of the second one is easy, since \(z=4\) it is just \(y\)
yashar806
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ok
yashar806
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then we add them togther?
anonymous
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now we use the fact that \(x+y=8\) and so \(y=8-x\)
anonymous
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now we add them together, because now we have an expression for the volume in one variable
it will be
\[V(x)=\frac{x^2}{4}+8-x\]
anonymous
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now if this is a calculus question you can take the derivative, set it equal to zero and solve
if not, you can say the vertex is at \(-\frac{b}{2a}\) with \(a=\frac{1}{4}\) and \(b=-1\)
yashar806
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got it
anonymous
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ok good
this has a min at the vertex, and i guess a max at the end point of the interval, so be careful about what that is
anonymous
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i guess it is \([0,8]\) since \(x\) must be in there somewhere
yashar806
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ok, thanks alot `
yashar806
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Hi just quick question, why we need interval there? how did you know that there should be an inteval?