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Minimizing Surface Area of a Can

  1. Jun 15, 2007 #1
    This is what i have so far

    SA = 2(pi)r^2 + 2(pi)r(245.45 / (pi)r^2)

    Derivative of SA = - 490.9r^2 + 4(pi)r

    0 = -490.9r^2 + 4(pi)r

    and i'm stuck at this step.. i've tried a bunch of things to try and solve for r but i can't seam to get a logical answer for the radius of the minimized pop can
  2. jcsd
  3. Jun 15, 2007 #2
    minimise, thats something new, well to minimise why just not have a can at all? make r=0. this is one of the roots of the equation you gave me and it it the only min turnning point as d^2/dx^2 at the point is greater than 0. Moreso the other root 0.025 has d^2y/dx^2 less than 0 making it a maximum turnning point. The working follows:

    0 = -490.9r^2 + 4(pi)r
    hence r=0 or 490.9r=4(pi)
    r=0 or r=0.02559
  4. Jun 15, 2007 #3
    You didn't differentiate S correctly. According to your post,

    [tex]S=2\pi r^{2} + 2\pi r(\frac{245.45}{\pi r^{2}}) = 2\pi r^{2} + \frac{490.9}{r}[/tex]


    [tex]\frac{dS}{dr}=4\pi r - 490.9r^{2}[/tex]

    Do you see where your error is?
  5. Jun 15, 2007 #4
    -490.9r^-2 not ^2
  6. Jun 15, 2007 #5

    thanks alot both of you lol .. it's funny how the simple mistakes often cause the most grief
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