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Difficult Precalc Questions

  1. Apr 3, 2008 #1
    Alright so currently I'm learning about using zeros of first and second derivites to find relative max and mins in an equation. That stuff is pretty simply. Also we're learning to differentiate equations in relation to a variable that is not in the problem. But, I cannot figure out these problems and would appreciate help.

    1. The problem statement, all variables and given/known data
    1. A cylindrical can of radius r inches and height h inches is to contain 36 cubic inches of liquid. The volume of the can is V= (pi)r^2h cubic inches and the total surface area is S = area of the sides + area of top and bottom = 2(pi)rh + 2(pi)r^2 square inches. The material for the top and the bottom of the can costs 22 cents per square inch and the material for the sides costs 18 cents per square inch. Find the values of r and h that minimize the cost of constructing the can.

    I started by differentiating V=(pi)r^2h in respect to dh/dsa and dr/sa I came up with dr/dsa = (r/-2h)*(dh/dsa). I don't really know where to go from here though.

    2. A Natural gas pipe is to be laid from the main gas line to the island airport located on the shoreline of an island as shown in the figure. the island is 2.2 miles from the mainland and the main gas line is 8 miles away from a point that is directly landward of the island. The cost of the pipe is 2.5 times times as great in the water as on land. At what point should the pipe meet the mainland shore in order to minimize total costs?

    island
    |
    2.2 miles
    |
    Land---------8 miles--------main line

    I used the pythagereon therom to determine the length of the third side (8.2970) but I don't know how to create a diffirential equation relating price in so I can find the minimum using it's derivitive.

    3. The number n, in thousands, of a certain type of radial tire that is sold each month and the monthly expenses a, in thousands of dollars, for advertising that tire are related by the equation .15a-9n+an+.02an^2-8.13=0. A tire is manufcaturer currently selling 3.5 thousand tires each month and spending $6000 per month on advertising. If plans call for increasing the advertising expenses by .4 thousand per month, how quickly should sales increase.

    sorry, I don't know where to begin on this problem.

    Any help would be greatly appreciated. Thanks for your time.
     
  2. jcsd
  3. Apr 3, 2008 #2

    cepheid

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    For the gas problem. Let the point at which the pipe meets the mainland be a variable (x). This means that it is distance of x from the directly landward point, and 8 - x from the mainline.

    Let the piping cost 'a' dollars per mile on land and 2.5a dollars per mile in the water.

    The distance d traversed in water is given by the pythagorea theorem

    [tex] d = \sqrt{x^2 + (2.2)^2} [/tex]

    Can you see that the cost function is:

    [tex] C(x) = 2.5ad + a(8-x) [/tex]

    ?
     
  4. Apr 3, 2008 #3

    cepheid

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    For the first problem, if the volume is constrained, then r is given in terms of h. (A cylinder of volume 36 cubic inches is only possible with certain height given the radius, or with a certain radius given the height.)

    So now this is a one-variable problem, and you can minimize the surface area with respect to r (or h, depending on which variable you got rid of using the constraint).
     
  5. Apr 3, 2008 #4
    Thanks for your help, so for the gas problem I can just replace the D in the cost formula and then find the zero of the derivitive to find the min value. I'll try that. Also, I'm still kind of confused about the cylindrical can problem. I don't think that the radius and the height are related like that. For the formula V=(pi) * r^2 * h there are going to be more than one possibility for r and h that will give a V of 34, right?
     
  6. Apr 4, 2008 #5

    cepheid

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    Sure there'll be more than one possibility, but that doesn't contradict anything I was trying to say.

    I meant that the radius and height will always be related as follows:

    [tex] V = \pi r^2 h [/tex]

    [tex] h = \frac{V}{\pi r^2} [/tex]

    [tex] V = \textrm{const.} [/tex]

    So any time you see h, you can replace it with:

    [tex] \frac{V}{\pi r^2} [/tex]

    Now you only have ONE variable (r). Yes, it's still a variable (there's more than one possibility that will lead to the same volume).

    Alternatively, you could have written everything in terms of h. It doesn't matter.
     
  7. Apr 4, 2008 #6

    cepheid

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    Yes, for the gas problem, writing d in terms of x in the cost formula and differentiating is exactly what I would do.
     
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