Question

Optimization! factory plans to manufacture cylindrical cans to hold 10 L of oil, where a litter is 1 dm^3. Find the dimensions that minimize the amount of metal needed to produce such a can.

Answer 1

@Hero

Answer 2

so, you want to keep the volume 10L while minimizing the surface area.

Answer 3

\(Volume~ \pi r^2h\\Surface~area:~2 \pi r^2+2\pi rh \)

Answer 4

okay! i'm not really sure what to do after though..

Answer 5

we know the volume is 10, so\[10=\pi r^2h\] but we have two unknowns, so we need to make it so that we have only 1 unknown.

Answer 6

so we are going to rewrite the h to become a function of r, so that we can substitude the h in the surface equation

Answer 7

so, writing \(10=\pi r^2h ~as~h=???\) what would h be? (in terms of r)

Answer 8

sqrt(10/hpi)

Answer 9

? not sure what you mean

Answer 10

or h=10/pir^2

Answer 11

hmm, you rewrote r as a function of h, that will probably work as well.

Answer 12

wait i got them mixed up now lol

Answer 13

\[r=\sqrt{\frac{ 10 }{ h \pi }}\]\[h=\frac{ 10 }{ \pi r^2 }\] either works.

Answer 14

now use \[Surface~area:~2 \pi r^2+2\pi rh\] and replace either the h or the r so that you have a function with either ONLY r (and pi and numbers ofc.) or ONLY h in it. You'll be able to solve it then.

Once you know the value of either r or h, the other one will be no problem. (or you can just do it once for both r and h)

Answer 15

how would I solve that though? sorry I have no clue what to do our teacher hasn't taught us this yet and she wants us to try to teach ourselves optimization before she teaches us

Answer 16

are you well versed with differentiation?

Answer 17

not really to be honestt

Answer 18

if we take \(\large h=\frac{ 10 }{ \pi r^2 }\) and instert that into \(Surface~area:~2 \pi r^2+2\pi rh\) then it becomes:

\(\large Surface~area:~2 \pi r^2+2\pi r\frac{ 10 }{ \pi r^2 }\) which we can simplify to
\(\large Surface~area:~2 \pi r^2+\frac{ 20 }{ r }\)

we could take the visual approach and draw a graph of it (either with graphic calculator or a n estimate by hand)

it would look something like this (only considering the positive side for obvious reasons)



Obviously, reading it from the graph can be handy when you have a calculator or want a rough estimate, but it's more precise (and sometimes even quicker) to just do it algebraically. Luckily, there's a reasonably easy way if you know how to take a derivative.

I don't really think it's my task to explain derivatives in full detail, however, the basics are like this:

the derivative f'(x) also known as \(\large \frac{ dy }{ dx }\) of a function f(x) is the slope of that function at a certain point. So, when the derivative f'(x) is 2, then the \(\underline {slope}\) of function f(x) is 2 at that exact point. Since we want to find an extremeity, the slope at that very point is 0 for just that point. (it is negative just before that point and it is positive just after that point)

therefore, if we find the derivative of the function and then solve the derivative for zero, we would at least know it's an extreme point (either maximum or minimum or corner) in this case we already know it's the minimum because this function does not have a maximum nor a corner.

The derivative of a simple function like this can be found by the following rule:

\[\Large ax^n \Rightarrow anx^{ n-1}\]

rewriting your function:

\( \large Surface~area:~2 \pi r^2+ 20r^{-1} \Rightarrow 4\pi r-20r^{-2} \)

rewriting to most simple form again

\[\large \frac{ dy }{ dr }=4\pi r - \frac{20}{r^2}\]

now solve for 0

\[\large 4\pi r - \frac{20}{r^2}=0\]
should give you the radius. Once you know the radius in decimeters, remember the cylinder has to contain 10 liters, so you can easily calculate the appropriate height as well.