- #1
Jordan_
- 20
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Alright, here is another problem I'm having trouble with. I used a technique I got off of someone on my other thread, and am just looking to see what should be done next.
Alright so first I got my equation for the surface area of a cylinder:
SA = 2(pi)rh + 2(pi)r^2
and converted 500mL to 0.0005m^3
Then I put in my cost numbers:
SA = 1.2[2(pi)r^2] + 0.4[2(pi)rh]
Now at this stage I used what I learned in the other thread and got the 'h' out of there by switching the volume formula around (V = (pi)r^2h) and substituting it for 'h'.
SA = (1.2)2(pi)r^2 + (0.4)2(pi)r[.0005/(pi)r^2)
Now I think I have to simplify and then take the derivative, but when I do that I get dr/dx's and then I get confused and don't know where to go. Anyone give me some tips?
A soup manufacturer wants to sell it's soup in 500mL cans. The metal for the top and bottom costs $1.20/m^2. The metal for the side costs $0.40/m^2. After the circles for the top and bottom are cut out of rectangle, the remaining metal will be scrapped. Find the dimensions of the can that will minimize the cost of materal.
ANS: 2.54m, 24.68m
Alright so first I got my equation for the surface area of a cylinder:
SA = 2(pi)rh + 2(pi)r^2
and converted 500mL to 0.0005m^3
Then I put in my cost numbers:
SA = 1.2[2(pi)r^2] + 0.4[2(pi)rh]
Now at this stage I used what I learned in the other thread and got the 'h' out of there by switching the volume formula around (V = (pi)r^2h) and substituting it for 'h'.
SA = (1.2)2(pi)r^2 + (0.4)2(pi)r[.0005/(pi)r^2)
Now I think I have to simplify and then take the derivative, but when I do that I get dr/dx's and then I get confused and don't know where to go. Anyone give me some tips?