## dessyj1 one year ago I need help visualizing this question. (Calc 1)

1. anonymous

Type le question

2. dessyj1

there was supposed to be an attachment

3. dessyj1

A metal through with equal semicircluar ends and open top is to have a capacity of 128pi cubic feet. Determine its radius r and length h is the trough is to require the least material for its construction.

4. dessyj1

trough*

5. triciaal

|dw:1433651448727:dw|

6. dessyj1

so the ends of the are semi-circle are closed?

7. triciaal

|dw:1433651580013:dw|

8. anonymous

Now just write an expression for the volume, Then one for the Surface area. Use the volume formula to turn the surface area formula into a single variable. Then Differentiate the last formula, set it equal to zero and solve. This will give you either h or r (depending on how you get on) Then substitute back into first formula to find the other variable.

9. triciaal

I see a cylinder split in half then turned on the side

10. dessyj1

I see a cylinder split in half too.

11. dessyj1

volume of a cylinder/2= 128pi?

12. dessyj1

then 2h+4r+2pi(r)=perimeter of the trough?

13. zepdrix

perimeter? hmm we're looking to minimize surface area, ya? :) remember how to find surface area of a cylinder?

14. dessyj1

the materials would be sum the dimensions of all the sides would it not?

15. zepdrix

no, that would take care of the .. edges i suppose, not the surfaces though.

16. zepdrix

|dw:1433648747657:dw|For a cylinder, we have two of these panels, one on top and one on bottom, ya? area of a circle is pi r^2. So to get the surface area of the top and bottom combined, we have 2pi r^2

17. dessyj1

SA of a cylinder= 2pi(r)h+ 2pi(r)^2

18. zepdrix

Or you can jump right to your formula :) yah that's fine hehe.

19. triciaal

|dw:1433652364767:dw|

20. zepdrix

Our trough will have exactly half of that surface area, so the formula we'll use is:$\Large\rm A=\pi r^2+\pi r h$I divided everything by 2 to cut it in half.

21. dessyj1

alright

22. dessyj1

the cylinder volume formula would have to be cut in half as well?

23. zepdrix

You're trying to minimize the area function. So eventually you want to get $$\Large\rm A'$$ and set it equal to zero to look for critical points. But first, you want A in terms of ONE VARIABLE so it's easier to work with.

24. zepdrix

Yes, tric put the 1/2 in front of the pi r^2 h to show that :)

25. zepdrix

the Area formula is what we're trying to minimize, the volume formula is our constraint. It relates r to h. It allows us to make a substitution in our Area formula. Do you see how that will work? :)

26. dessyj1

yup

27. dessyj1

would isolate for h in the volume formula then sub it into the SA formula

28. zepdrix

good good good

29. dessyj1

then derive it.

30. dessyj1

$SA \prime =4Pir-(512\Pi/r ^{2})?$

31. dessyj1

r=$\sqrt[3]{128}$

32. zepdrix

Mmm that's what I'm coming up with also! yay

33. dessyj1

okay, so when i originally tried to figure out the perimeter i was trying to get just the sides?

34. dessyj1

|dw:1433649816855:dw|

35. zepdrix

um um um :) ya perimeter gives us the distance across all of those edges i guess.

36. dessyj1

Okay, thank you so much.

37. zepdrix

Were you able to find the corresponding height using that radial value? :)

38. zepdrix

Remember that they want the radius and height that minimize the area :D

39. dessyj1

yes h=10.07 inches

40. dessyj1

or

41. dessyj1

$h= 256\div \sqrt[3]{128}^2$

42. zepdrix

cool :) that will simplify a lil bit if you want both your height and radius in exact value,$\Large\rm h=\frac{256}{(128)^{2/3}}=\frac{2\cdot128}{(128)^{2/3}}=2(128)^{1/3}$