Maximum Volume: 72cm^2 of sheet steel is used for a closed container with a square ba

[limaecho1988]

I am hoping somebody can give me a clue where to start with this one.

I have question where I need calculate the maximum volume of can. The details given below:

72cm^2 of sheet steel is used for a closed container with a square base. If width of the base is w cm, show that the Volume Vcm^2(I assume a typo?) so maybe V cm^3 is given by:

V = w (18 - (w^2/2))

Seem to be getting a bit confused with the shape of the thing. I figure the surface area for the base would be w^2.

I understand using derivatives to find and classify maximum and minimum values but can't seem to get past the first hurdle on this one.

BRgds,

 

[HallsofIvy]

There are a lot of ambiguities here. It says, for example, that this is to be made from a "72 cm^2 sheet of steel". One way (but only "one way"). typical in problems like this, is, if we have a square sheet of steel, we cut from the sides into the square that is to be the bottom, then fold up the sides, giving a box (with a w by w base but no top). If we do that, then, since a square of area 72=8*9= 2(4*9) square cm must have each side of length \(\displaystyle 6\sqrt{2}\) cm. Cutting from the side to the w by w square leaves a side length of \(\displaystyle 3\sqrt{2}- w\). Bending that up we have volume, \(\displaystyle V= w^2(3\sqrt{2}- w)= 3\sqrt{2}x^2- w^3\).

However, I think that what you are doing is to just cut up the 72 square cm sheet into any convenient pieces and, where necessary, weld them together to make the box. Assuming base w by w cm and height h, we have one w by w piece and 4 w by h pieces for a total are of w^2+ 4wh= 72. We can solve that for h= (72- w^2)/4w= 18/w- w/4. Then the volume is V= w^2h= w^2(18/w- w/4)= 18w- w^3/4.

 

[MarkFL]

The surface area \(\displaystyle S\) of the box (whose height is \(\displaystyle h\)) in \(\displaystyle \text{cm}^2\) will be:

\(\displaystyle 2w^2+4wh=72\)

Solve for \(\displaystyle h\):

\(\displaystyle \displaystyle h=\frac{18}{w}-\frac{w}{2}\)

Thus, the volume of the box is given by:

\(\displaystyle V=w^2h=?\)

 

[MarkFL]

There are a lot of ambiguities here...

Yes, it took me quite some time to realize that we aren't taking a sheet and removing rectangles to make a shape that can be folded into a closed box.

 

[limaecho1988]

Thanks once again.