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Help With Set up

  1. Feb 7, 2005 #1
    If a cube is expanding so the surface area is increasing at a constant rate of 7 in^2/sec How fast is the volume increasing at the instant the surface area is 204 in^2 ?

    How would I go about starting this problem to set it up?

    surface area = 6x^2
    volume = x^3
  2. jcsd
  3. Feb 7, 2005 #2


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    Use the chain rule.

    You're given:

    You have to find:
  4. Feb 7, 2005 #3
    [tex] \frac{dS}{dx}\frac{dx}{dt} [/tex] = 7

    [tex] 12 x \frac{dx}{dt}=7 [/tex]

    [tex] \frac{dx}{dt}=\frac{7}{12x} [/tex]

    [tex] \frac{dV}{dt}=\frac{dV}{dx} \frac{dx}{dt} [/tex]

    [tex] \frac{dV}{dt}= 3x^2 \frac{7}{12x} = \frac {7x}{4} [/tex]

    [tex] SA = 204 = 6x^2 [/tex]

    [tex] SA = \sqrt{34} [/tex]

    [tex] rate = \frac{7\sqrt{34}}{4} [/tex]

    [tex] rate= 10.2 in^3/sec [/tex]
    Last edited: Feb 7, 2005
  5. Feb 7, 2005 #4
    Does my work seem correct?
  6. Feb 7, 2005 #5


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    you mean
    [tex] x= \sqrt{34}[/tex]

  7. Feb 7, 2005 #6
    Thank you for your help

    So the general formula for something like this would simply be
    SA rate= X
    Volume rate wanted at Surface Area = Y

    [tex] \frac{X \sqrt{\frac{Y}{6}}}{4} = Volume Rate [/tex]
  8. Feb 7, 2005 #7


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    Hold on,this is a cube we're talking about.So if your formula works for the cube,i can guantee u it doesn't work for any other closed surface.

  9. Feb 7, 2005 #8
    Yes I figured it was for the cube only, thanks for the clafication and the help though
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