## yashar806 Group Title I need help, please one year ago one year ago

1. yashar806

I have attached the question

2. satellite73

we can do this

3. yashar806

yes, could you explain it ?

4. satellite73

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

5. satellite73

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

6. yashar806

why is it 1/4 ?

7. satellite73

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

8. yashar806

ok

9. satellite73

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

10. yashar806

ok?

11. satellite73

the volume of the second one is easy, since $$z=4$$ it is just $$y$$

12. yashar806

ok

13. yashar806

14. satellite73

now we use the fact that $$x+y=8$$ and so $$y=8-x$$

15. satellite73

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

16. satellite73

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 $$b=-1$$

17. yashar806

got it

18. satellite73

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

19. satellite73

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

20. yashar806

ok, thanks alot `

21. yashar806

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