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yashar806 Group TitleBest ResponseYou've already chosen the best response.0
I have attached the question
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
we can do this
 one year ago

yashar806 Group TitleBest ResponseYou've already chosen the best response.0
yes, could you explain it ?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
ok first off z is 4, so lets forget z and replace it by 4
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
the volume of the first tube, the one with side \(x\) is \(4\times (\frac{1}{4}x)^2\)
 one year ago

yashar806 Group TitleBest ResponseYou've already chosen the best response.0
why is it 1/4 ?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
because your entire length is \(x\) and you are folding it in to a square so the side of the square is \(\frac{1}{4}x\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
and therefore the area of the square is \((\frac{1}{4}x)^2=\frac{x^2}{16}\) and since \(z=4\) the volume is \(4\times \frac{x^2}{16}=\frac{x^2}{4}\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
the volume of the second one is easy, since \(z=4\) it is just \(y\)
 one year ago

yashar806 Group TitleBest ResponseYou've already chosen the best response.0
then we add them togther?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
now we use the fact that \(x+y=8\) and so \(y=8x\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
now we add them together, because now we have an expression for the volume in one variable it will be \[V(x)=\frac{x^2}{4}+8x\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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 \(b=1\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i guess it is \([0,8]\) since \(x\) must be in there somewhere
 one year ago

yashar806 Group TitleBest ResponseYou've already chosen the best response.0
ok, thanks alot `
 one year ago

yashar806 Group TitleBest ResponseYou've already chosen the best response.0
Hi just quick question, why we need interval there? how did you know that there should be an inteval?
 one year ago
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