Question

I need help, please

Answer 1

I have attached the question

Answer 2

we can do this

Answer 3

yes, could you explain it ?

Answer 4

ok first off z is 4, so lets forget z and replace it by 4

Answer 5

the volume of the first tube, the one with side \(x\) is \(4\times (\frac{1}{4}x)^2\)

Answer 6

why is it 1/4 ?

Answer 7

because your entire length is \(x\) and you are folding it in to a square
so the side of the square is \(\frac{1}{4}x\)

Answer 8

ok

Answer 9

and therefore the area of the square is \((\frac{1}{4}x)^2=\frac{x^2}{16}\) and since \(z=4\) the volume is \(4\times \frac{x^2}{16}=\frac{x^2}{4}\)

Answer 10

ok?

Answer 11

the volume of the second one is easy, since \(z=4\) it is just \(y\)

Answer 12

ok

Answer 13

then we add them togther?

Answer 14

now we use the fact that \(x+y=8\) and so \(y=8-x\)

Answer 15

now we add them together, because now we have an expression for the volume in one variable
it will be
\[V(x)=\frac{x^2}{4}+8-x\]

Answer 16

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 \(b=-1\)

Answer 17

got it

Answer 18

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

Answer 19

i guess it is \([0,8]\) since \(x\) must be in there somewhere

Answer 20

ok, thanks alot `

Answer 21

Hi just quick question, why we need interval there? how did you know that there should be an inteval?