Question

(Optimization) Using the same amount of aluminum, a soft drink company is trying to design a
cylindrical can that can hold the maximum amount of drink. What should be the base diameter to
height ratio of such a can to yield the maximum capacity? (plz show work)

Answer 1

Same amount of aluminum means that the surface area is a restriction that you must meet when optimizing the volume.
Are any numerical values given?

Answer 2

no numerical values are given

Answer 3

Alrighty then, I assume the question wants the general solution. A is your constant surface area.
\[A = 2 \pi r^2 + 2 \pi r h\]\[V = \pi r^2 h\]If we solve for h is the surface area equation and plug it into the volume, we get,\[h = \frac {A - 2 \pi r^2}{2 \pi r}\]\[V = \pi r^2 (\frac {A - 2 \pi r^2}{2 \pi r}) = \frac {A r}{2} - \pi r^3\]

Answer 4

Now just differentiate the volume and find the optimal value for the radius. Then use that to find the height. They should be in terms of A.
Can you do that?

Answer 5

id appreciate if u could do the whole thing out i hav no idea how to do it...

Answer 6

\[V' = \frac {A}{2} - 3 \pi r^2\]Set that to zero and solve for when the volume is a maximum.\[V' = 0 = \frac {A}{2} - 3 \pi r^2\]Then volume will be maximized when \[r = \sqrt{ \frac {A}{6 \pi}}\]Substitute back in to get the maximum height.\[h = \frac {A}{2 \pi r} - r = \frac {A}{ 2 \pi \sqrt{ \frac {A}{6 \pi}}} - \sqrt{\frac {A}{6 \pi}}\]Is that right so far? Its a lot of messy stuff, I need verification.

Answer 7

i think it is from my notes

Answer 8

tho i cant really understand my notes so ...

Answer 9

Well, is the algebra right?

Answer 10

precalculus

Answer 11

:o Your doing this in precalculus? Then there must be a simpler solution than this way... Anyways, let me just finish up this problem...\[d = 2r = 2 \sqrt{ \frac {A}{6 \pi}}\]\[h = \frac {A}{ 2 \pi \sqrt{ \frac {A}{6 \pi}}} - \sqrt{\frac {A}{6 \pi}}\]The diameter to height ratio is then...\[\frac {d}{h} = \frac {2 \sqrt{ \frac {A}{6 \pi}}} {\frac {A}{ 2 \pi \sqrt{ \frac {A}{6 \pi}}} - \sqrt{\frac {A}{6 \pi}}} = 1\]All that ugly stuff just simplifies down to 1. The amount of aluminum used will be optimized when the height and diameter of the can are the same.

Answer 12

Sorry if you can't understand it :( I did it via calculus. Sometimes calculus isn't the simplest way to go about doing things...