Question

create a cylindrical can that minimizes cost of materials the can must contain 110 cubic inches of soup. The top and bottom of the can cost $0.0014 per square
inch, while the sides cost only $0.0007 per square inch. What are the dimensions of the
can that minimize the cost of materials?

I keep getting .0000007 for the radius. I know something must be wrong

Answer 1

im getting r = 2.061 inch, whats ur cost equation ?

Answer 2

I am using a really long equation. ((2r*pi)*110/(2r*pi))*.0007)+((r^2*pi*2)*.0014)

Answer 3

It made sense when I wrote it down. it doesnt make sense when I graph it

Answer 4

ok, Volume = 110

\(\pi r^2h = 110 \\ h = \frac{110}{\pi r^2} ---------- (1)\)

Answer 5

so I am missing the part where I get the correct volume, right?

Answer 6

\(Area = 2\pi r^2 + 2\pi r h\)

\(= 2\pi r^2 + 2\pi r \frac{110}{\pi r^2}\)

\(= 2\pi r^2 + 2 \frac{110}{ r} -------------(2)\)

Answer 7

hmm didnt get you, you have the same equations as mine, so far ?

Answer 8

the cost function become :

\(Cost = 2\pi r^2(0.0014) + \frac{220}{r} (0.0007) ----------(3)\)

Answer 9

now minimize the \(Cost\)

Answer 10

what is the 220/r?

Answer 11

\(Cost\) function is equation \((2)\), multiplied by costs

Answer 12

in equation \((2)\), i wrote 2.110/r, in equation \((3)\), i wrote it as 220/r
i thought u would get it by nw as u r working on this for a while.

Answer 13

oh. sorry, I just didnt catch where it came from

Answer 14

ok np :) see if u can complete the rest of the prob

Answer 15

I really dont know what to do next. and thats only the first out of 5 on my homework. hopefully the rest go better

Answer 16

if u understood this problem, the rest would go better, as we spent so much time on this :)

Answer 17

I understand the concept and all. And I put a lot of thought into my mess of an equation. something is just not clicking right now

Answer 18

its okay :) happens sometimes wid me, good to understood the basic concept.

so next we just minimize the \(Cost\)

Answer 19

1) Find the derivative of \((3)\)
2) equate it to 0
3) find the value of \(r\)

Answer 20

I tried graphing the new equation. but its still turning out the same. its a parabola ith the minimum right next to 0

Answer 21

*With the minimum

Answer 22

this problem is from calculus, right ?

Answer 23

I know we can restrict the domain to positives. I just dont know how I am not shifting the problem to the right at all

correct, this is a calc problem

Answer 24

then do it the calculus way, i have listed the steps to be carried out above.

Answer 25

like i know it cant be 0, that is impossible for a comtainer, so the parabola must be moved to the right by some amount. I am just missing that part or something

Answer 26

if you zoomin, you would see graph is going minimum at around .1 or something... its not 0 for sure. but again lets not graph it right away, first do it the calculus way.

Answer 27

thats the graph wolfram is giving.