Thread: Minimize Area of Soup Can (Involves Quotients)

1. Minimize Area of Soup Can (Involves Quotients)

I answered the question my way and I have 0.0298 m for the optimal radius of the can... Now I know the book's answer is wrong because it says 21m*16m or some silly thing.

Here's the question:

A soup manufacturer wants to sell its soups in 500-ml cans. The metal for the top is $1.20/m^2 The metal for the sides costs$0.40/m^2. After the circles for the top and bottom are cut out of a rectangle, the remaining metal is scrapped. Find the dimensions of the can that will minimize material cost.

So I have a few equations I used:
pi*((r)^2)*h=.0005 m^3

and

to minimize cost:
C=2*pi*(r^2)*1.20 + 0.40*(2pi(r)h)

2. You are asked for "the dimensions of the can", which will be two values: the radius and the height. You seem to have arrived at only one value...?

Since 1 mL = 1cc, the volume (in terms of linear units) is 500 cc. Since the costs are given in terms of square meters, though, we'll have to convert. Since 100 cm = 1 m, then 10 000 cm^2 = 1 m^2 and 1 000 000 cc = 1 m^3. Then 500 cc = 0.0005 m^3.

The relevant equations are:

. . . . . $\mbox{volume: }\, V\, =\, \pi r^2 h \,=\, \frac{1}{2000}$

. . . . . $\mbox{surface area: }\, SA\, =\,2 \pi r^2\, +\, 2 \pi r h$

. . . . . $\mbox{cost: }\, C\, =\, (1.2)(2 \pi r^2)\, +\, (0.4)(2 \pi r h)\, =\, 2.4 \pi r^2\, +\, 0.8 \pi r h$

We can solve the "volume" equation for h in terms of r:

. . . . . $h\, =\, \frac{1}{2000 \pi r^2}$

Then the "cost" function is:

. . . . . $C\, =\, 2.4 \pi r^2\, +\, \frac{0.8 \pi r}{2000 \pi r^2}\, =\, 2.4 \pi r^2\, +\, \frac{1}{250r}$

Differentiating with respect to "r" gives:

. . . . . $C'\, =\, 4.8 \pi r \, -\, \frac{1}{250r^2}$

Setting equal to "zero" gives:

. . . . . $4.8 \pi r \, =\, \frac{1}{250r^2}$

. . . . . $1200 \pi r^3\, =\, 1$

. . . . . $r^3\, =\, \frac{1}{1200 \pi}$

I don't get your answer or the book's, but I've probably gone wrong somewhere...

3. you get very close to mine. I had r^3=1/12000pi so one of us messed up somewhere. I'll have to go back and check, but at the same time, my answer with the given radius was fairly close to reasonable (a 6cm diameter can).
I only posted ym one answer because I wanted to see if my radius is correct.... Getting the height after that is a joke.

Yes, in your math, you missed a zero near the beginning.... .8/2000=1/2500 you put 1/250

4. Area and volume

No-one has yet got it right.
Originally Posted by mike_302
...After the circles for the top and bottom are cut out of a rectangle, the remaining metal is scrapped.
The area used for the top and bottom has nothing to do with $\pi$, has it? It's simply $8r^2$.