Finding Maximums and Minimums for Shapes: making tanks from sheets

[karendeeley]

Hi everyone; posted this in Calculus because of the maximum/minimum part, but perhaps it should have gone in geometry.

I have looked at this question from many different angles. I am unsure how I would be able to find three different unknown values, given the amount of information provided. From what I can gather, only one sheet of aluminium can be used for the fuel tank.
Thanks for any help.

------------
A sheet metal fabrication company produces rectangular aluminium sheets that will be used to make fuel tanks for small boats. The sheets are 120 cm by 80cm. The company has employed you to design the fuel tank. The fuel tank can be either a rectangular prism or a cylinder.
1. Find the dimensions of the rectangular prism tank that will maximise its volume. Determine the capacity of the tank in litres.

 

[stapel]

Hi everyone; posted this in Calculus because of the maximum/minimum part, but perhaps it should have gone in geometry.

I have looked at this question from many different angles. I am unsure how I would be able to find three different unknown values, given the amount of information provided. From what I can gather, only one sheet of aluminium can be used for the fuel tank.
Thanks for any help.

------------
A sheet metal fabrication company produces rectangular aluminium sheets that will be used to make fuel tanks for small boats. The sheets are 120 cm by 80cm. The company has employed you to design the fuel tank. The fuel tank can be either a rectangular prism or a cylinder.
1. Find the dimensions of the rectangular prism tank that will maximise its volume. Determine the capacity of the tank in litres.

Are you allowed to assume that the entire sheet will be used, or are you supposed to cut parts away and fold what remains to create the shape? I suspect that you're supposed to do the latter. If so, then:

Let the sheet be cut, so that you end up with something that looks like this:

cut sheet:

   *--------*    EDIT:
   |        |d   <==IGNORE THIS PICTURE
   |        |    <==SEE BELOW FOR CORRECTION
*--*--------*--*      ||    ||
|  |        |  |      ||    ||
|  |   L    | h|      \/    \/
*--*--------*--*
   |        |
   |        |
   *--------*
   |        |
   |        |
   *--------*

...where "h" is the height, "d" is the (front-to-back) depth, and "L" is the (side-to-side) length.

Clearly, 4d is equal to one of the sheet's original dimensions, and L + 2h is equal to the other dimension. Let's assume that the horizontal in the above is the 80 and the vertical is 120. Then we have 4d = 120, so d = 30. Also, we have L + 2h = 80. The volume will be given by:

. . . . .\(\displaystyle V\, =\, 30Lh\)

This has two variables, but we can solve the equation "L + 2h = 80" for one of the variables, and substitute. Where does this lead? (And what do you get if you reverse the assumptions about the horizontal and the vertical?)

---

Edit: The above drawing is indeed incorrect. Sorry! Here's a correction to this post:

cut sheet:

   *--------*
 h |   L   h| h <==shorter
*--*--------*--*
|  |       d|  | <==taller
|  |   L    | h|
*--*--------*--*
  h|   L   h| <==shorter
   *--------*
  d|        |
   |   L   d| <==taller
   *--------*

---

Thanks to karendeeley for checking my work and finding that something was wrong; thanks to mmm4444bot for calling out and correcting my error! :cool:

 

[karendeeley]

Thanks, stapel, for getting back so fast.
I do still have a question, then, though. You said that 4d = 120, so d = 30. But for this to work, then H would also be 30, for the D and H to line up as an edge on the shape. And if H is 30, then L = 20. This forms a very skinny shape. Is there a way to work out of D is definitely 1 value, 30, or if it's got two different values, one of which equals the new H value? For example, perhaps two of the long rectangles (L x D) are 20 x 30, and two are 20 x 40. Otherwise, D is a constant, and forms a square cross-section with each side being 1D; for this to line up with the end face, with a sidelength of H, then H also has to be 30. This makes a very small tank.

 

[mmm4444bot]

I do still have a question, then, though. You said that 4d = 120, so d = 30. But for this to work, then H would also be 30, for the D and H to line up as an edge on the shape.

I may have misunderstood stapel's approach, or her diagram may have a typo. The flaps folded up to form the walls all need to have the same height.

If we cut out h by h squares at the top, then we have this:

[FONT=courier new]        L
   *--------*
   |       h|
*--*--------*--*
|  |D       | h| 
|  |    L   |  | D
*--*--------*--*
   |        |h
   *--------*
   |        |
  D|        |
   *--------*
        L[/FONT]

This tank has a base measuring L by D, and the height is h.

Volume = L*D*h

120 = 2*h + 2*D

80 = L + 2*h

Rewriting L and h in terms of D, I worked out the maximum volume to be 33802 cubic centimeters (rounded).

However, since a square has the largest rectangular area with a fixed perimeter, I think a larger volume is possible by building a tank with square ends (i.e., D=h). What do you think? :cool:

 

[stapel]

I may have misunderstood stapel's approach, or her diagram may have a typo. The flaps folded up to form the walls all need to have the same height.

You're quite right; I made a typo in my pretty picture, and then was led astray.... shame on me!

The picture is correct, as mmm4444bot posted it:

If we cut out h by h squares at the top, then we have this:

[FONT=courier new]        L
   *--------*
   |       h|
*--*--------*--*
|  |D       | h| 
|  |    L   |  | D
*--*--------*--*
   |        |h
   *--------*
   |        |
  D|        |
   *--------*
        L[/FONT]

Going across (and assuming that this is 80), we then have:

. . . . .80 = h + L + h = L + 2h

Going down (and assuming that this is 120), we also have:

. . . . .120 = h + D + h + D = 2h + 2D

These equations can be restated as:

. . . . .80 - 2h = L

. . . . .60 = h + D

The volume equation is always the same:

. . . . .V = LhD

Because two of the variables (namely, L and D) can be stated in terms of a number and the third variable (namely, h), the equations are better rearranged as:

. . . . .\(\displaystyle 40\, -\, \dfrac{L}{2}\, =\, h\)

. . . . .\(\displaystyle 60\, -\, D\, =\, h\)

Using this, we can solve for one of L and D in terms of the other. For instance:

. . . . .\(\displaystyle 40\, -\, \dfrac{L}{2}\, =\, 60\, -\, D\)

. . . . .\(\displaystyle 80\, -\, L\, =\, 120\, -\, 2D\)

. . . . .\(\displaystyle 2D\, -\, 40\, =\, L\)

Then the volume equation, V, can be stated in terms of just the one variable, D:

. . . . .\(\displaystyle V(D)\, =\, (2D\, -\, 40)(60\, -\, D)(D)\)

Differentiate, etc, etc. :wink: