Question

A can of peanuts has a circular base and top, and it is cylindrical in shape. The volume is
500 cm3. If the can is to require the least amount of material, what must its dimensions be?
Give your answer correct to one decimal place.

Answer 1

what do you get for r ?

Answer 2

What have you considered on this problem so far?

Answer 3

The volume is given. If you know the formula for volume, you can set up one relationship between the variables of radius and height.
Then, to minimize the amount of materials, the only relevant part of the can using material is the surface itself, so we are minimizing surface area essentially.

Answer 4

i used the area formula after ? is that wrong ?

Answer 5

What area formula did you use?

Answer 6

2pirh+2pir^2

Answer 7

That looks good to me. Then we just need to use the first volume equation to eliminate one variable.

Answer 8

so this is where i am stuck r^3=250/pi

Answer 9

\( V = \pi r^2 h \) This is the volume of a cylinder.
\( 500 = \pi r^2 h \) We need to solve for one of these variables to use in surface area, which allows us to minimize surface area in terms of one dimension.

Answer 10

yep i did this h=500pir^2

Answer 11

500/pir^2 i mean

Answer 12

So you substituted that into surface area:
\(S = 2 \pi r^2 + 2 \pi r h \) \( h = \dfrac{500}{\pi r^2} \)
\(S = 2 \pi r^2 + 2 \pi r \left( \dfrac{500}{\pi r^2} \right) \)

Answer 13

With this equation, we can use optimization to find the value for r that minimizes the surface area.

Answer 14

yea so i was stuck at r^3=250/pi

Answer 15

\( S = 2 \pi r^2 + 2\cancel{\pi r } \dfrac{500}{\cancel{\pi r} r} \)
\( S = 2 \pi r^2 + \dfrac{1000}{r} \)
\( S' = 4 \pi r - \dfrac{1000}{r^2} = 0 \)
\( 4 \pi r = \dfrac{1000}{r^2} \)
\( r^3 = \dfrac{1000}{4 \pi } \)
You would take the cube root from here, or power to 1/3.
\( \sqrt[3]{r^3} = \sqrt[3]{\dfrac{250}{\pi}} \)
That makes sense?

Answer 16

okey yea then next step ??

Answer 17

i know we sub it in

Answer 18

That gives us the radius. Substituting it into the volume formula, then:
\( 500 / \left[ \pi \left( \sqrt[3]{\dfrac{250}{\pi}} \right) ^2\right] = h \)

Answer 19

I presume "dimensions of the can" implies diameter and height. So we could double the radius for diameter.

Answer 20

what do u get after that step ??

Answer 21

do u get 12.599 for h ?

Answer 22

I obtained h = 8.602 according to calculator.

Answer 23

wait what did u do ?

Answer 24

\( 500 / \left[ \pi \left( \sqrt[3]{\dfrac{250}{\pi}} \right) ^2\right] = h \)
This has to be done on a calculator, being careful of the order in which items are attended to.
500/(pi*(250/pi)^(2/3))

Answer 25

i canclced the pi thats what i did wrong i think

Answer 26

Yeah, those two pi's don't have the same degree since one is raised to the 2/3 power. The main simplification would be to cancel 2/3 from both leaving \( \pi^{1/3} \), but it is usually easier to just enter it all in a calculator to avoid those mess-ups.