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Math Help - Minimizing Cost

  1. #1
    Member ~berserk's Avatar
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    Minimizing Cost

    An observatory is to be in the form of a right circular cylinder surmounted by a hemispherical dome. If the hemispherical dome costs 8 times as much per square foot as the cylindrical wall what are the most economic dimensions for a volume of 6000 cubic feet?
    The radius of the cylindrical base (and of the hemisphere)= ?
    The height of the cylindrical base=?

    So to start this I have the equation:here

    sorry for having to link it, Latex isn't working for me and neither is image upload.
    But I just need to know where to go from here
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  2. #2
    MHF Contributor

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    Yes, taking r to be the radius of the cylinder (and so the radius of the hemispherical top), the total volume is \pi r^2h+ (2/3)\pi r^3= 6000.

    Now, you want to minimize the cost so you need to find a formula for the cost in terms of r and h. The surface area of the curved side of the cylinder is the area of a rectangle of height h and length 2\pi r (the circumference of the circular base), 2\pi r h. The surface area the hemispherical cap is 2\pi r^2. We are only concerned with "relative" cost so take the cost of 1 unit of area for the cylinder wall to be "1" and then the cost of 1 unit of area of the dome is "8". The total (relative) cost would be 2\pi rh+ 16\pi r^2.

    Yes, one way to do this would be to replace h with the formula you have so that you have a formula for cost in r only. Set the derivative of that equal to 0 to find the optimal r.

    Another way to do this would be to use "Lagrange multipliers" if you know that method. Minimize 2\pi rh+ 16\pi r^2 subject to the constraint \pi r^2h+ 2\pi r^3/3= 6000.
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