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Easy Application of Calc Question

  1. May 17, 2012 #1
    but I have a headache.

    Here's the question:
    A laundry powder company wishes to produce a 27mm^3 washing powder capsule in the shape of a rectangular box. Due to production rescrictions, the width and the depth of the capsule must be equal lengths, and all dimensions of the capsule must be equal to at least 2mm. To ensure the rate at which the capsule is dissolved during washing, it is required that the surface areas of the capsule is as large as possible. What should the dimensions of the capsule be so that its surface area is maximised.


    I've got that if the width and depth = a, and the length = b, the volume = b*a^2
    Therefore, 27= b*a^2, but where do I go from here.
     
    Last edited: May 17, 2012
  2. jcsd
  3. May 17, 2012 #2
    Use the fact that the volume must be 27mm³ to find b in terms of a. Then find the surface area in terms of a and b, and then plug in the previously found formula for b in order to find the surface area in terms of a. Use calculus to find the maximum of this function subject to the constraints that a≥2 and b≥2. Then dance a jig. The jig is very important -- solving math problems feels much better if you dance afterward than if you do not.:smile:
     
  4. May 17, 2012 #3
    Thanks for the response. So I've done that.

    b=27/a^2 and a=27/b

    The equation for the total surface area in terms of a and b is:
    Area = 2a^2 + 4ab

    subbing b=27/a^2 gives:
    Area = 2a^2 + 108/a

    dA/da = 4a - 108/a^2

    The area equation is a weird graph where the function is always increasing after a local minimum at (3,54). You'd expect to equate at a local max for a maximum area question like this. How can apply this to the question?
     
  5. May 18, 2012 #4
    Very good, except that you're missing a square root in your formula for a (this does not affect your subsequent calculations). Now, since you've learned that the only place where the derivative is zero is a local minimum, that means that the maximum must occur at the boundary. So either a=2 or 27/a^2=2. Which gives you the greater area?
     
  6. May 18, 2012 #5
    Woops. So a=√27/√b?

    So your saying the domain of A is 2<=a<=27/a^2? Can you explain why to solve 27/a^2=2?

    Anyway,

    A(2)=62mm^2

    So, for the second part, should I solve 27/a^2=2 for a, then sub this value the Area equation?

    The above equation solved for a=√27/√2.

    A(√27/√2)=12√6 + 27
    which = 56.39mm^2

    Have I done the right thing?
     
    Last edited: May 18, 2012
  7. May 18, 2012 #6
    Yes, that's pretty much it. The reason the right endpoint of the domain is √(27/2) is because this corresponds to b=2. Anyway, as you see from your calculations, the area is greater when a=2, so that will be where the maximal surface area occurs.
     
  8. May 18, 2012 #7
    Thanks for your help.
     
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